Question

find the value of-
[-1/4a+1/2b-1]^2

Answer 1

Step-by-step explanation:

Answer:

\begin{gathered}Expansion\:of \:(\frac{1}{4a}-\frac{1}{2b}+1)^{2}\\=\frac{1}{16a^{2}}+\frac{1}{4b^{2}}+1\\-\frac{1}{4ab}-\frac{1}{b}+\frac{1}{2a}\end{gathered}

Expansionof(

4a

1

−

2b

1

+1)

2

=

16a

2

1

+

4b

2

1

+1

−

4ab

1

−

b

1

+

2a

1

Step-by-step explanation:

Expansion\:of \:(\frac{1}{4a}-\frac{1}{2b}+1)^{2}Expansionof(

4a

1

−

2b

1

+1)

2

/* we know the algebraic identity:

$$\begin{gathered}\boxed {(x+y+z)^{2}\\=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx}\end{gathered}$$ *\

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

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

=$$\begin{gathered}\frac{1}{16a^{2}}+\frac{1}{4b^{2}}+1\\-\frac{1}{4ab}-\frac{1}{b}+\frac{1}{2a}\end{gathered}$$