Question

$\huge\bold\red{{HELP}}$​

Answer 1

Solution:

$\boldsymbol{A) \: \dfrac{x^{2}y}{2xy^{2}}}$

This can be written as:

$\implies \sf{\dfrac{(x)(x)(y) }{2(x)(y)(y)}}$

On cancelling the common terms,

$\implies \sf{\dfrac{(\not{x})(x)(\not{y}) }{2(\not{x})(\not{y})(y)}}$

$\implies \sf{\dfrac{x}{2y}}$

$\therefore \boxed{\bf{\dfrac{x^{2}y}{2xy^{2}}=\dfrac{x}{2y}}}$

$\boldsymbol{B) \: \: \dfrac{4q-1}{1-4q}}$

This can also be written as:

$\implies \sf{\dfrac{4q-1}{-(4q-1)}}$

Cancelling (4q - 1) in numerator and denominator,

$\implies \sf{\dfrac{1}{-1}}$

$\implies \sf{-1}$

$\therefore \boxed{\bf{\dfrac{4q-1}{1-4q}=-1}}$

$\boldsymbol{a. \: \: \dfrac{-3a^{3}b^{3}}{6ab^{3}}}$

This can be simplified and written as:

$\implies \sf{\dfrac{3(-1)(a)(a)(a)(b^{3})}{3(2)(a)(b^{3})}}$

On cancelling the common terms,

$\implies \sf{\dfrac{\not{3}(-1)(\not{a})(a)(a)(\not{b^{3}})}{\not{3}(2)(\not{a})(\not{b^{3}})}}$

$\implies \sf{\dfrac{-a^{2}}{2}}$

$\therefore \: \boxed{\bf{\dfrac{-3a^{3}b^{3}}{6ab^{3}}=\dfrac{-a^{2}}{2}}}$

$\boldsymbol{b. \: \: \dfrac{4x-2}{8x^{2}-4}}$

Taking 2 as a common factor,

$\implies \sf{\dfrac{2(2x-1)}{2(4x^{2}-2)}}$

$\implies \sf{\dfrac{2x-1}{4x^{2}-2}}$

$\therefore \: \boxed{\bf{\dfrac{4x-2}{8x^{2}-4}=\dfrac{2x-1}{4x^{2}-2}}}$

Answer 2

$\huge\green{Solution}$

Question :- A

Also Written as :-

$\frac{x \times x \times y}{2 \times x \times y \times y} = \frac{x}{2y}$

Question:-B

Also written as:-

$\frac{ \: 4q - 1}{ - (4q - 1)} = - 1$

Try it a:-

Also written as:-

$\frac{3 \times - 1 \times a \times a \times a \times b \times b \times b}{3 \times 2 \times a \times b \times b \times b} = \frac{ - {a}^{2} }{2}$

Try it b:-

$\frac{4x - 2}{8 {x - 4}^{2} } = \frac{2(2x - 1)}{2( {4x }^{2} - 2 ) } = \frac{2x - 1}{4 {x}^{2} - 2 }$

$\huge\red{Thank \: You}$