Bài 8:

a) \(x^2-2x+1=25\)

\(\Rightarrow\left(x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=5^2\\\left(x-1\right)^2=\left(-5\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

b) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Rightarrow25x^2+10x+1-25x^2+9=30\)

\(\Rightarrow10x+10=30\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

c) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Rightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Rightarrow x^3-1-x^3+4x=5\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\frac{3}{2}\)

d) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=15\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Rightarrow24x+25=15\)

\(\Rightarrow24x=-10\)

\(\Rightarrow x=\frac{-5}{12}\)

Bài 9:

a) \(-x^2+6x-15=-x^2+6x-9-6=-\left(x^2-6x+9\right)-6=-\left(x-3\right)^2-6\)

Ta có: \(\left(x-3\right)^2\ge0\Rightarrow-\left(x-3\right)^2\le0\Rightarrow-\left(x-3\right)^2-6\le-6\)

\(\Rightarrow-x^2+6x-15\) luôn âm với mọi \(x\)

b) \(-9x^2+24x-18=-9x^2+24x-16-2=-\left(9x^2-24x+16\right)-2=-\left(3x-4\right)^2-2\)

Ta có: \(\left(3x-4\right)^2\ge0\Rightarrow-\left(3x-4\right)^2\le0\Rightarrow-\left(3x-4\right)^2-2\le2\)

\(\Rightarrow-9x^2+24x-18\) luôn âm với mọi \(x\)

c) \(\left(x-3\right)\left(1-x\right)-2=x\left(1-x\right)-3\left(1-x\right)-2=x-x^2-3+3x-2=-x^2+4x-5\)

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

Ta có: \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-1\le-1\)

\(\Rightarrow\left(x-3\right)\left(1-x\right)-2\) luôn âm với mọi x

d) \(\left(x+4\right)\left(2-x\right)-10=x\left(2-x\right)+4\left(2-x\right)-10=2x-x^2+8-4x-10=-x^2-2x-2\)

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

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow-\left(x+1\right)^2\le0\Rightarrow-\left(x+1\right)^2-1\le-1\)

\(\Rightarrow\left(x+4\right)\left(2-x\right)-10\) luôn âm với mọi \(x\)

16x²+24x+9

=(4x)²+2.4x.3+3²

=(4x+3)²

bài 6:

a)=(x+2)(x^2-2x+4)

b)=(3-2x)(9+6y+4y^2)

c)=(y+1)(y^2+y+1)

B6:

d)=(4x-1/2 y)(16x^2+2xy+1/4y^2)

e)=(5x^2-3x^3)(25x^4+15x^4+9^6)

mik sửa lại câu 

c)=(y^2+1)(y^4-y+1)

(2xy+1)²-(2x+y)²

=(2xy+1-2x-y)(2xy+1+2x+y)

=[(2xy-y)-(2x-1)][(2xy+y)+(2x+1)]

=[y(2x-1)-(2x-1)][y(2x+1)+(2x+1)]

=(2x-1)(y-1)(2x+1)(y+1)