\(=\frac{x^4-x^2-3x^2+3}{x^4-x^2+7x^2-7}=\frac{x^2\left(x^2-1\right)-3\left(x^2-1\right)}{x^2\left(x^2-1\right)+7\left(x^2-1\right)}=\frac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\frac{x^2-3}{x^2+7}\)

HELP ME

\(A\left(x\right)=\dfrac{4x^4+81}{2x^2-6x+9}\)

\(=\dfrac{4x^4+36x^2+81-36x^2}{2x^2-6x+9}\)

\(=\dfrac{\left(2x^2+9\right)^2-\left(6x\right)^2}{2x^2+9-6x}\)

\(=\dfrac{\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)}{2x^2+9-6x}\)

\(=2x^2+6x+9\)

=>\(M\left(x\right)=2x^2+6x+9\)

\(=2\left(x^2+3x+\dfrac{9}{2}\right)\)

\(=2\left(x^2+3x+\dfrac{9}{4}+\dfrac{9}{4}\right)\)

\(=2\left(x+\dfrac{3}{2}\right)^2+\dfrac{9}{2}>=\dfrac{9}{2}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{2}=0\)

=>\(x=-\dfrac{3}{2}\)

>=9/2 là sao vậy

a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)

b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)

d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)

D . x² + 4x + 4 = ( x + 2 )²

Câu D tui ghi sai rồi xin lỗi nha

D)x²+5x+6\x²+4x+4

(-3).8/8.6 rút gọn

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)

c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)

d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)