1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)

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

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

2: \(x^2-8x+7=0\)

=>\(x^2-x-7x+7=0\)

=>\(x\left(x-1\right)-7\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

1:

a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)

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

\(=x^2+1\)

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

a)
\(=\left(x+2y\right)\left(x^2-xy+y^2\right)-3xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+y^2-3xy\right)\)
\(=\left(x+2y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+2\right)\left(x-2\right)^2\)
b)
\(3x\left(2x-1\right)\left(2x+1\right)=0\)
3x=0 =>x=0
hoặc 2x-1=0 => 2x=1=>x=1/2
hoặc 2x+1=0=>2x=-1=>x=-1/2

bạn giải giúp mình một câu ở dưới nữa được không?

\(a) x^3-4x^2+8x-32=(x^3-4x^2)+(8x-32)=x^2(x-4)+8(x-4)=(x^2+8)(x-4)​\)
th1 \(X^2+8\)=0

      \(X^2=-8( vô lí)\)

Th2 x-4=0

        X=4

Phương trình có tập nghiệm S=4

Ta có: \(x^3-4x^2+8x-32=0\)

\(\Leftrightarrow x^2\left(x-4\right)+8\left(x-4\right)=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

Ta có: \(7x^2-28=0\)

\(\Leftrightarrow7\left(x^2-4\right)=0\)

\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-2\right\}\)