Question

Using suitable identity expand (a÷4-b÷2+1)whole square

Answer 1

Hi!
The answer of your question is as follows:
using identity (a+b+c)^2 we get;
(a/4)^2+(-b/2)^2+(1)^2+2×a/4×-b/2+2×-b/2×1+2×1×a/4
a^2/16+b^2/4+1-ab/4-b+a/2
HOPE that this will help you.........
Please mark it as brainliest...........

Answer 2

Answer:

The expanded form is $(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}$        

Step-by-step explanation:

Given : Expression $(\frac{a}{4}-\frac{b}{2}+1)^2$

To find : The expression value using identity?

Solution :

Expression $(\frac{a}{4}-\frac{b}{2}+1)^2$

Using identity,

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

Where, $a=\frac{a}{4}, b=-\frac{b}{2}, c=1$

Substitute the value,

$(\frac{a}{4}-\frac{b}{2}+1)^2=(\frac{a}{4})^2+(-\frac{b}{2})^2+1^2+2((\frac{a}{4})(-\frac{b}{2})+(-\frac{b}{2})(1)+(1)(\frac{a}{4}))$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+2(\frac{-ab}{8}-\frac{b}{2}+\frac{a}{4})$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+2(\frac{-ab-4b+2a}{8})$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+\frac{-ab-4b+2a}{4}$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2+16+4(-ab-4b+2a)}{16}$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2+16-4ab+16b+8a}{16}$

$(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}$

Therefore, The expanded form is $(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}$