\(a,4x^2-y^2+2y-1\)

\(=4x^2-\left(y^2-2y+1\right)\)

\(=\left(2x\right)^2-\left(y-1\right)^2\)

\(=\left(2x-y+1\right)\left(2x+y-1\right)\)

câu 2 :

       \(100+10x+\frac{1}{4x^2}\)

<=> \(\left(10\right)^2+2.10.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2\)

<=> \(\left(10+\frac{1}{2}x\right)^2\)

câu 1 :

        \(\frac{25}{x^2}-\frac{5}{2}+\frac{x^2}{16}\)

\(\Leftrightarrow\)\(\left(\frac{5}{x}\right)^2-2.\frac{5}{x}.\frac{x}{4}+\left(\frac{x}{4}\right)^2\)

\(\Leftrightarrow\)\(\left(\frac{5}{x}+\frac{x}{4}\right)^2\) k mình nhé bạn !!

Đề sai r bn nhé

\(4xy+4x^2y^2+1\)

\(=4x^2y^2+4xy+1\)

\(=\left(2xy\right)^2+2.2xy.1+1^2\)

\(=\left(2xy+1\right)^2\)

nhân tung

(xy² - 1/2)(2 + 4xy²)

= 4(xy² - 1/2)(xy² + 1/2)

= 4[(xy²)² - (1/2)²]

= 4(x²y⁴ - 1/4)

=2x-1

\(\left(4x-3\right)\left(3x+2\right)-\left(6x+1\right)\left(2x+5\right)+1\)

\(=\left(12x^2-9x+8x-6\right)-\left(12x^2+2x+30x+5\right)+1\)

\(=\left(-x-32x\right)+\left(-6-5+1\right)=-33x-10\)

(2x-1)(2x+1)(4x² +1)

= (4x² - 1)(4x²+1)

=16x⁴ - 1 

(2x-1).(2x+1).(4x^2+1)

= (4x^2 - 1).(4x^2+1)

=  (4x^2)^2  -  1

=  16x^4  - 1

Ta có:4x²-(2x-3)(2x+3)= 9x

          4x²-4x²-6x+6x+9=9x

          9=9x

           x=1

Vậy x=1