Calculations and problems: corrected math exercises in 2nd grade.

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Answer key for math exercises in 2nd grade on calculations and problems. Develop skills with fractions and powers.

Exercise 1:

Either $X=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}$

a. Show that

first of all $3-2\sqrt{2}<3+2\sqrt{2}$

Let be the function $f:x\,\mapsto \,\sqrt{x}$ , this function is strictly increasing on $%5B0;+\infty%5B$ .

so $0<3-2\sqrt{2}<3+2\sqrt{2}$

$\Rightarrow$ $f(0)<f(3-2\sqrt{2})<f(3+2\sqrt{2})$

thus

$\sqrt{0}<\sqrt{3-2\sqrt{2}}<\sqrt{3+2\sqrt{2}}$

and therefore

$\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}<0$

b. Calculate .

$X^2=(\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}})^2=(\sqrt{3-2\sqrt{2}})^2-2\sqrt{3-2\sqrt{2}}\sqrt{3+2\sqrt{2}}+(\sqrt{3+2\sqrt{2}})^2\\=3-2\sqrt{2}-2\sqrt{(3-2\sqrt{2})(3+2\sqrt{2})}+3+2\sqrt{2}$

$\\=6-2\sqrt{9-(2\sqrt{2})^2}$

$\\=6-2\sqrt{9-8}$

$\\=4$

c. Deduce the value of There are two possible values X=2 or X=-2

from 1 X<0 , we deduce that X = – 2 .

Exercise 2:

Calculate:

$A=(-2)^3\times \,5+3^2\times \,2^4-5\times \,2^2$

$B=9\times \,(\frac{2}{3})^2-(3^2\times \,2)^4-5\times \,2^2$

$B=9\times \,\frac{4}{9}-3^8\times \,2^4-20$

Exercise 3:

$A=1+\frac{3}{4}\times \,\frac{2}{3}-\frac{1}{6}:\frac{3}{4}$

$A=1+\frac{2}{4}-\frac{1}{6}\times \,\frac{4}{3}$

$A=1+\frac{1}{2}-\frac{2}{9}$

$A=\frac{18}{18}+\frac{9}{18}-\frac{4}{18}$

${\color{DarkRed}\,A=\frac{23}{18}}$

$B=(7-\frac{3}{2})\times \,(\frac{25}{7}+\frac{3}{5})$

$B=(\frac{14}{2}-\frac{3}{2})\times \,(\frac{25\times \,5}{7\times \,5}+\frac{3\times \,7}{5\times \,7})$

$B=\frac{11}{2}\times \,(\frac{25\times \,5}{35}+\frac{3\times \,7}{35})$

$B=\frac{11}{2}\times \,(\frac{125}{35}+\frac{21}{35})$

$B=\frac{11}{2}\times \,(\frac{146}{35})$

$B=\frac{11}{2}\times \,(\frac{2\times \,73}{35})$

$B=\frac{11\times \,73}{35}$

$B=\frac{803}{35}$

Exercise 4:

1. Develop and reduce:

a= (5x+1)(2x+3)

${\color{red}\,a=10x^2+17x+3}$

b= (4x-5)(7x-1)

${\color{DarkRed}\,b=28x^2-39x+5}$

c= (2x+5)(7x-3)

${\color{DarkRed}\,c=14x^2+29x-15}$

d= (-4x-6)(2x-1)

${\color{DarkRed}\,d=-8x^2-8x+6}$

2. Expand and reduce:

e= (5x+1)(2x+3)+(5x+1)(x+2)

${\color{DarkRed}\,e=15x^2+28x+5}$

f= (4x-5)(7x-1)-(4x-5)(3x+2)

${\color{DarkRed}\,f=16x^2-32x+15}$

g= (-4x-6)(2x-1)+(2x-3)(8x-11)

${\color{DarkRed}\,g=8x^2-54x+39}$

h= (x-8)(5+3x)-(x-8)(7-x)

${\color{DarkRed}\,h=4x^2-34x+16}$

Exercise 5:

Frame when $-\sqrt{5}\leq\,\,x<\,1$ .

then

${\color{DarkRed}\,0\leq\,\,x^2\leq\,\,5}$

Exercise 6:

Factor the following expressions:

$A=(3x-5)(3x-5)+(3x-5)\times ,1$

Exercise 7:

$A=2\sqrt{3}-4\sqrt{3}+\sqrt{27}$

$A=2\sqrt{3}-4\sqrt{3}+\sqrt{9\times ,3}$

$A=2\sqrt{3}-4\sqrt{3}+3\sqrt{3}$

${\color{DarkRed},A=\sqrt{3}}$

$B=\frac{3-\sqrt{6}}{4-\sqrt{5}}$

$B=\frac{,(3-\sqrt{6},,)\times ,,(4+\sqrt{5},,),}{,(4-\sqrt{5},,)\times ,,(4+\sqrt{5},,)}$

$B=\frac{,(3-\sqrt{6},,)\times ,,(4+\sqrt{5},,),}{4^2-\sqrt{5}^2}$

$B=\frac{12+3\sqrt{5}-4\sqrt{6}-\sqrt{30},}{16-5}$

$B=\frac{12+3\sqrt{5}-4\sqrt{6}-\sqrt{30},}{11}$

Exercise 8:
1. In what form is an even number written ?

$2k\,(k\in\mathbb{R}).$

2. In what form is an odd number written?

$2k+1\,(k\in\mathbb{R}).$

3. Show that the square of an even number is an even number.

$(2k)^2=4k^2=2\times \,(2k^2)=2K$ with $K=2k^2\in\mathbb{R}$

Conclusion: the square of an even number is an even number.

Exercise 9:
1. Calculate the sum of 5 consecutive integers.

It’s up to you to do some tests.

2. Show that the sum of five consecutive integers is a multiple of 5.

Let x be the smallest integer .

Either

$x+(x+1)+(x+2)+(x+3)+(x+4)=5k\,(k\in\mathbb{N})$

So the sum of 5 consecutive integers is a multiple of 5.

Exercise 10:
1. Calculate the product of four consecutive integers and add 1.

What do we notice? (Make several attempts)
2x3x4x5+1=121=11²
5x6x7x8+1=1681=41²
6x7x8x9+1=3025=552

2. Show that, for any real x, we have :

$x(x\,+\,1)(x\,+\,2)(x\,+\,3)\,+\,1\,=\,(x^2\,+\,3x\,+\,1)^2$

Explain the result observed in question 1.

The product of 4 consecutive numbers plus 1 can be written as the square of a number.

Exercise 11:
a. Decompose into the product of prime numbers 220 and 284.

$220=2^2\times \,5\times \,11$

$284=2^2\times \,71$

b. Check that 220 and 284 are friendly.

The divisors of 220 are 1;2;4;5;10;11;20;22;44;55;110;220.

The divisors of 284 are 1;2;4;71;142;284

or let’s calculate the sum of the divisors except m :

and

Conclusion: 220 and 284 are two friendly integers.

Exercise 12:

Write in interval form $(x\in...)$ :

$-5<x\leq\,\,2\,:\,x\in%5D-5;2%5D\\x\geq\,\,\frac{3}{2}\,:\,%5B\frac{3}{2};+\infty%5B\\x\leq\,\,-\frac{1}{4}\,:\,%5D-\infty;-\frac{1}{4}%5D\\x>-5\,et\,x\leq\,\,3,5\,:\,%5D-5;3,5%5D$

Exercise 13:

1. Hervé must factor A.

Here’s his answer:

${\color{Blue}\,A=(2x+5)(2x+5+x-4)}$

${\color{Blue}\,A=(2x+5)(3x+1)}$

Test the equality obtained by Hervé for

What can we conclude from this?

$A=(2\times \,0+5)(3\times \,0+1)=15$

and if we replace 0 in the original expression, we obtain :

$A=(-2\times \,0+5)^2+(2\times \,0+5)(0-4)+2\times \,0+5=25-20+5=10$

Conclusion : Hervé made a mistake when factoring.

2. To factor A, we can think of writing :

Then factor A correctly.

$A=(2x+5)%5B\,2x+5\,+\,x-4\,+1%5D$

${\color{DarkRed}\,A=(2x+5)(\,3x+2)}$

Exercise 14:

Factor each expression by highlighting a common factor.

$A=9a+15=3\times \,3a+3\times \,5=3(3a+5)\\B=3x^2-15x=3x\times \,x-3x\times \,5=3x(x-5)\\C=8x-x^2(5x-1)=x(8-x(5x-1))\\D=(3x-2)^2-(2x-1)(3x-2)\\=(3x-2)%5B(3x-2)-(2x-1)%5D\\=(3x-2)(3x-2-2x+1)=(3x-2)(x-1)$

Exercise 15:

$A=(7x+1)^2=49x^2+14x+1\\B=(x-3)^2=x^2-6x+9\\C=(-3-2x)^2=9+12x+4x^2\\D=(5x-4)^2=25x^2-40x+16\\E=(3x+1)(3x-1)=9x^2-1$

Exercise 16:

$E=\sqrt{(\sqrt{2}-\sqrt{5})^2}-\sqrt{(2\sqrt{2}-\sqrt{5})^2}$

$E=\,%7C\,\sqrt{2}-\sqrt{5}\,\,%7C-\,%7C\,2\sqrt{2}-\sqrt{5}\,\,%7C$

$E=\sqrt{5}\,-\sqrt{2}-(2\sqrt{2}-\sqrt{5})$

$E=\sqrt{5}\,-\sqrt{2}-2\sqrt{2}+\sqrt{5}$

$E=\sqrt{5}\,-3\sqrt{2}+\sqrt{5}$

$E=2\sqrt{5}\,-3\sqrt{2}$

Exercise 17:
$x=\,%7C\,\sqrt{2}-1\,\,%7C=\sqrt{2}-1\,car\,\sqrt{2}>1.$

$y=\,%7C\,\sqrt{3}-5\,\,%7C=5-\sqrt{3}\,car\,5>\sqrt{3}.$

$z=\,%7C\,\pi-5\,\,%7C=5-\pi\,car5>\pi.$

$t=\,%7C\,7-2\pi\,\,%7C=7-2\pi\,car\,7>2\pi.$

$v=\,%7C\,3-\pi\,\,%7C=\pi-3\,car\,\pi>3.$

Exercise 18:

$%5B-6;2%5B\cap\,%5D-4;1%5D=%5D-4;1%5D$

$%5D-1;3%5D\cap\,%5B2;4%5B=%5B2;3%5D$

$%5D-\infty;4%5B\cap\,%5D2;+\infty%5B=%5D2;4%5B$

$%5D-\infty;-3%5D\cap\,%5B2;+\infty%5B=\,\emptyset$

$%5B-1;+\infty%5B\cap\,%5B3;+\infty%5B=%5B3;+\infty%5B$

$%5D-\infty;2%5D\cap\,%5B2;4%5B=\,\{\,2\,\,\}$

Exercise 19:

$A=\frac{a^{-4}b^5(ac^2)^3}{(ba^{-2})^5}$

$A=\frac{a^{-4}b^5a^3c^6}{b^5a^{-10}}$

$A=\frac{a^{-4}a^3c^6}{a^{-10}}$

$A=\frac{a^{-1}c^6}{a^{-10}}$

$A=a^{-1+10}c^6$

${\color{DarkRed}\,A=a^{9}c^6}$

Exercise 20:

The golden number is the number $\phi\,=\frac{1+\sqrt{5}}{2}$ .

Check the following equalities:

1. $\phi^2\,=\phi+1$ .

$\phi\,^2-\phi\,-1=\,(\,\frac{1+\sqrt{5}}{2}\,\,)^2-\,(\frac{1+\sqrt{5}}{2}\,\,)-1=\frac{1+2\sqrt{5}+5}{4}-\frac{2+2\sqrt{5}}{4}-\frac{4}{4}=\frac{1+2\sqrt{5}+5-2-2\sqrt{5}-4}{4}=\frac{0}{4}=0$

2. $\phi\,=\frac{1}{\phi}+1$ .

Let’s multiply by $\phi$ :

$\phi^2=\frac{1}{\phi}\times \,\phi+\phi$

$\phi^2=1+\phi$

$\phi^2=\phi\,+1$ (this is the equality of question 1.)

3. $\phi^3\,=2\phi+1$

We know that :

$\phi^2=\phi\,+1$

Let’s multiply this equality by $\phi$ :

$\phi^3=\phi^2+\phi$

$\phi^3=\phi+1+\phi$ (because $\phi^2=\phi+1$ )

$\phi^3=2\phi+1$

Exercise 21:
The speed of light is estimated at $3\times \,10^8$ m/s

and the average distance Earth-Sun at 149 million kilometers.

Calculate the time needed for a light signal from the Earth to reach the
Sun.

$t=\frac{d}{v}=\frac{149\times \,10^6\times \,10^3}{3\times \,10^8}$

$t=\frac{149\times \,10^9}{3\times \,10^8}$

$t=\frac{149\times \,10^1}{3}$

${\color{DarkRed}\,t\simeq\,497\,s.}$

Exercise 22:

For $x=-\frac{1}{2}$ , calculate :

$A=4x^3-2x^2+x+3=4(-\frac{1}{2})^3-2(-\frac{1}{2})^2+(-\frac{1}{2})+3$

$A=-\frac{4}{8}-\frac{2}{4}-\frac{1}{2}+3$

$A=-\frac{4}{8}-\frac{4}{8}-\frac{4}{8}+3$

$A=-\frac{12}{8}+\frac{24}{8}$

$A=\frac{12}{8}=\frac{3}{2}$

$B=\frac{x^3-1}{(x-1)(x^2+x+1)}$

$B=\frac{(-\frac{1}{2})^3-1}{((-\frac{1}{2})-1)((-\frac{1}{2})^2+(-\frac{1}{2})+1)}$

$B=\frac{(-\frac{1}{2})^3-1}{(-\frac{3}{2})(\frac{1}{4}-\frac{2}{4}+\frac{4}{4})}$

$B=\frac{-\frac{1}{8}-\frac{8}{8}}{(-\frac{3}{2})(\frac{1}{4}-\frac{2}{4}+\frac{4}{4})}$

$B=\frac{-\frac{9}{8}}{(-\frac{3}{2})(\frac{3}{4})}$

$B=\frac{-\frac{9}{8}}{-\frac{9}{8}}$

${\color{DarkBlue}\,B=1}$

Exercise 23:

Prove that the diagonal of a square of side is $a\sqrt{2}$ .

In the right-angled triangle ABC at B, according to the direct part of the Pythagorean theorem :

$AC=\sqrt{2a^2}$ (a length is positive or zero)

$AC=\sqrt{2}\times \,\sqrt{a^2}$

$AC=\sqrt{2}a$

${\color{DarkRed}\,AC=a\sqrt{2}}$

Exercise 24:

Exercise 25:

Exercise 26:

Exercise 28:
1. For each line, reconstruct the sentence using Si……then …. or …if and only if ….:

IF IT RAINS THEN I take my umbrella.

IF I in the middle of [AB] THEN AI=BI

$a\ge\,b$ IF AND ONLY IF $a-b\ge\,0$ .

IF $a\le\,3$ THEN $\,a\le\,5$ .

IF THEN ABC is isosceles at A.

Exercise 29:
For n natural numbers, compare the following numbers:

$\frac{n+1}{n+2}\,\le1\le\frac{n+6}{n+3}\le\frac{n+7}{n+3}$

Exercise 30:
Let $a\ge\,0\,;\,b\ge\,0$ ,compare the numbers :

Suppose that $\,\sqrt{a+b}\le\sqrt{a}+\sqrt{b}$
So $\,0\le(\sqrt{a+b})^2\le(\sqrt{a}+\sqrt{b})^2$
So $\,0\le(\sqrt{a+b})^2\le(\sqrt{a}+\sqrt{b})^2$
So $\,0\le\,a+b\le\,a+b+2\sqrt{a}\sqrt{b}$
So $\,0\le\sqrt{a}\sqrt{b}$
this being always true since $\,\sqrt{a}\ge0,\sqrt{b}\ge0$

so $\,\sqrt{a+b}\le\sqrt{a}+\sqrt{b}$

Exercise 31:
1. Complete using the symbols $\,\in\,;\notin$

a. $\sqrt{2}\in%5D1;3%5B$

b. $\frac{2}{\sqrt{2}}\,\in%5B\sqrt{2};5%5D$

2. Specify the interval corresponding to :

a. $\,%5B2;5%5D\cup\,%5D-1;7%5D=%5D-1;7%5D$

b. $%5B-1;\pi%5B\cup\,%5D\sqrt{2};5%5D=%5B-1;5%5D-\{\pi;2}$

c. $%5B3;+\infty%5B\cup\,%5D0;3%5B\cup\,\{3\}=%5D0;+\infty%5B$

Exercise 32:
a.

$\,\frac{3}{8}\times (3-\frac{1}{3})=\frac{3}{8}\times (\frac{9}{3}-\frac{1}{3})=\frac{3}{8}\times \frac{8}{3}=1$

$\frac{1+\frac{1}{2}}{2-\frac{23}{7}}\times (3-\frac{1}{3})=\frac{\frac{3}{2}}{-\frac{9}{7}}\times \frac{8}{3}=\frac{3}{2}\times (-\frac{9}{7})\times \frac{8}{3}=-\frac{36}{7}$

2. Simplify and then write the following fraction in scientific form:

$\frac{(6\times 10^{-2})^2\times \,3^2\times \,10^{-4}}{3^3\times \,10^{12}}=\frac{3^2\times 2^2\times \,3^2\times 10^{-4}\times \,10^{-4}}{3^3\times \,10^{12}}=3\times \,2^2\,\times \,10^{-8-12}\\=12\times \,10^{-20}$

3. Simplify the following entries:

a. $\,\sqrt{343}-10\sqrt{112}+\sqrt{7}=\sqrt{49\times \,7}-10\sqrt{16\times \,7}+\sqrt{7}=7\sqrt{7}-40\sqrt{7}+\sqrt{7}=-32\sqrt{7}$

b. $\,\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})\times \,(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})\times \,(\sqrt{3}-\sqrt{2})}=\frac{(\sqrt{3}-\sqrt{2})^2}{3-2}=3-2\times \sqrt{2}\times \sqrt{3}+2\\=5-2\sqrt{6}$

c. $\,(1-2\sqrt{2})^2=1-2\times 2\sqrt{2}+(2\sqrt{2})^2=1-4\sqrt{2}+8\\=9-4\sqrt{2}$

4. Give the prime factor product decomposition of the number $A=34\times \,12=2\times 17\times 2^2\times 3=2^3\times 3\times 17$ .

Exercise 33:
a. Indicate the nature of the following numbers:

$A=1+\frac{1}{3}$ : rational $B=\frac{\frac{5}{3}}{\frac{2}{9}}$ : rational

$C=\sqrt{7^{500}}$ : integer D=1+ $\pi$ : irrational

b. Simplify the writing of the following number:

$\,A=(3\sqrt{2}+5\sqrt{2})(3\sqrt{6}-2\sqrt{2})=8\sqrt{2}(3\sqrt{6}-2\sqrt{2})=24\sqrt{12}-16\times 2\\=24\sqrt{4\times 3}-32=48\sqrt{3}-32$

Answer key to the exercises on calculations and problems in 2nd grade.
After having consulted the answers to these exercises on calculations and problems in grade 2, you can return to the exercises ingrade 2.

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