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

\(=x^2+1\)

Ta có 3 x 2 + 8x + 5 = 0

ó 3 x 2 + 3x + 5x + 5 = 0 ó 3x(x + 1) + 5(x + 1) = 0

ó (3x + 5)(x + 1) = 0 

Vậy  x = - 5 3 ;   x = - 1

Đáp án cần chọn là: A

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

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

\(\Leftrightarrow x^2\left(x-1\right)^2-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)