Question

factorise 1/16x2+1/4y2+1-1/4xy-y+1/2x

Answer 1

$\textbf{Given:}$

$\mathsf{\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+1-\dfrac{1}{4xy}-y+\dfrac{1}{2x}}$

$\textbf{To factorize:}$

$\mathsf{\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+1-\dfrac{1}{4xy}-y+\dfrac{1}{2x}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+1-\dfrac{1}{4xy}-y+\dfrac{1}{2x}}$

$\textsf{This can be written as}$

$\mathsf{=\left(\dfrac{1}{4x}\right)^2+\left(\dfrac{1}{2y}\right)^2+1^2-2\left(\dfrac{1}{4x}\right)\left(\dfrac{1}{2y}\right)-2\left(\dfrac{1}{2y}\right)(1)+2(1)\left(\dfrac{1}{4x}\right)}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca}}$

$\mathsf{=\left(\dfrac{1}{4x}-\dfrac{1}{2y}+1\right)^2}$

$\implies\boxed{\mathsf{\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+1-\dfrac{1}{4xy}-y+\dfrac{1}{2x}=\left(\dfrac{1}{4x}-\dfrac{1}{2y}+1\right)\left(\dfrac{1}{4x}-\dfrac{1}{2y}+1\right)}}$

$\textbf{Find more:}$

1.X2-(a-1/a)x+1 factorise the following  

https://brainly.in/question/4301404#  

2.Factorise x^4-(x-z)^4​  

https://brainly.in/question/14993176#

Answer 2

Step-by-step explanation:

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