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Answer 1

Key point:-

Question 1, 2 uses linear equations. If we subtract the recurring digits of each other, we will be left with fractions.

In Question 4, 6, polynomial identity $(a+b)(a-b)=a^{2}-b^{2}$ is suitable in rationalization as we can square both terms.

In Question 5, we use the laws of exponents.

In Question 7 we use factor theorem.

In Question 8, 9, 11, we use $ma+mb=m(a+b)$.

In Question 12, we use $(a+b)^{2}=a^{2}+2ab+b^{2}$.

In Question 13, we use $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$.

In Question 14, 15, we use $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$.

Solution:-

Question 1.

Given: $x=1.\overline{36}$

$100x=136.\overline{36}\ \ \ \ \ (1)$

$x=1.\overline{36}\ \ \ \ \ (2)$

Subtracting $(2)$ from $(1)$ yields $99x=135$.

$\implies x=\dfrac{135}{99}$

$\implies \boxed{x=\dfrac{15}{11} }$

Question 2.

Given: $x=0.7\overline{32}$

$1000x=732.\overline{32}\ \ \ \ \ (1)$

$10x=7.\overline{32}\ \ \ \ \ (2)$

Subtracting $(2)$ from $(1)$ yields $990x=725$.

$\implies x=\dfrac{725}{990}$

$\implies \boxed{x=\dfrac{145}{198}}$

Question 3.

The calculation steps are as follows.

Given: $(\sqrt{3} +\sqrt{11} )^{2}(\sqrt{3} -\sqrt{11} )^{2}$

$=\{(\sqrt{3} )^{2}-(\sqrt{11} )^{2}\}^{2}$

$=(-8)^{2}$

$=\boxed{64}$

Question 4.

Simply let's multiply 1 to the fraction. If we multiply $\dfrac{8-3\sqrt{5} }{8-3\sqrt{5} }$, which is one, does not affect the value.

Given: $\dfrac{1}{8+3\sqrt{5} }$

$=\dfrac{1}{8+3\sqrt{5} }\times \dfrac{8-3\sqrt{5} }{8-3\sqrt{5} }$

$=\dfrac{8-3\sqrt{5} }{8^{2}-(3\sqrt{5} )^{2}}$

$=\boxed{\dfrac{8-3\sqrt{5} }{19} }$

Question 5.

Given: $243^{\frac{6}{5} }$

$=(3^{5})^{\frac{6}{5} }$

$=3^{5\times \frac{6}{5} }$

$=3^{6}$

$=\boxed{729}$

Question 6.

• $x=\dfrac{3-\sqrt{13} }{2}$

$\implies \dfrac{1}{x}=\dfrac{2}{3-\sqrt{13} }$

$\implies \dfrac{1}{x} =\dfrac{2(3+\sqrt{13} )}{-4}$

$\implies \dfrac{1}{x} =-\dfrac{3+\sqrt{13} }{2}$

$\implies \boxed{x+\dfrac{1}{x} =-\sqrt{13} }\ \ \ \ \ (1)$

Squaring both sides of the equation $(1)$

$\implies x^{2}+2+\dfrac{1}{x^{2}} =13$

$\implies x^{2}+\dfrac{1}{x^{2}} =\boxed{11}$

Question 7.

By factor theorem, we simply substitute $y=5$ into the polynomial.

Remainder: $5^{3}+5^{2}-2\cdot5+5$

$=125+25-10+5$

$=\boxed{145}$

Question 8.

Firstly the HCF of the two terms is $12x^{2}$.

Given: $48x^{3}-36x^{2}$

$=\boxed{12x^{2}(4x-3)}$

Question 9.

The HCF of the two terms is $5xy^{2}z$.

Given: $15x^{3}y^{2}z-25xy^{2}z^{3}$

$=5xy^{2}z(3x^{2}-5z^{2})$

Question 10.

This polynomial cannot be factorized.

Question 11.

First, we group then factorize two terms.

Given: $6ab-b^{2}+12ac-2bc$

$= b(6a-b)+2c(6a-b)$

$=\boxed{(6a-b)(b+2c)}$

Question 12.

We observe a difference of two complete squares.

Given: $(36a^{2}-12a+1)-(25b^{2})$

$=(6a-1)^{2}-(5b)^{2}$

$=\boxed{(6a+5b-1)(6a-5b-1)}$

Question 13.

We observe a difference of two complete cubes.

Given: $1-27x^{3}$

$=-(27x^{3}-1)$

$=\boxed{-(3x-1)(9x^{2}+3x+1)}$

Question 14.

It is a complete cube, as

$x^{3}-9x^{2}y+27xy^{2}-27y^{3}$

$=(x)^{3}+3(x)^{2}(-3y)+3(x)(-3y)^{2}+(-3y)^{3}$

$=\boxed{(x-3y)^{3}}$

Question 15.

Given: $8x^{3}-(2x-3y)^{3}$

$=8x^{3}-(8 x^3 - 36 x^2 y + 54 x y^2 - 27 y^3)$

$=\boxed{36x^{2}y-52xy^{2}+27y^{3}}$

Done!

Answer 2

Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. It is the building block for everything in our daily lives, including mobile devices, architecture (ancient and modern), art, money, engineering, and even sports

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