\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)

\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>x=1

vậy ...

\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)

\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

vậy ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

\(\left[{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=1\\x^2=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)

Giải:

a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)

\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)

\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)

\(\Leftrightarrow4x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

Vậy ...

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)

\(\Leftrightarrow30x-37=4\)

\(\Leftrightarrow30x=41\)

\(\Leftrightarrow x=\dfrac{41}{30}\)

Vậy ...

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)

\(\Leftrightarrow x^3+8-x^3-3=14x\)

\(\Leftrightarrow5=14x\)

\(\Leftrightarrow x=\dfrac{5}{14}\)

Vậy ...

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

\(\Leftrightarrow x^3+1-x^3-3x=2\)

\(\Leftrightarrow1-3x=2\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Vậy ...

a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)

=> \(x^2-2x-x-3x+7+9x=6\)

=> \(x^2-2x-x^2-3x+7+9x=6\)

=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)

=> \(4x=-1\)

Vậy \(x=\dfrac{-1}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)

=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)

=> \(42x=43\)

Vậy \(x=\dfrac{43}{42}\)

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)

=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)

=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)

=> \(5=14x\)

Vậy \(x=\dfrac{5}{14}\)

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

=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)

=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)

=> \(-3x=1\)

Vậy \(x=\dfrac{-1}{3}\)

b: \(\Leftrightarrow\left(3x-1\right)^2=25\)

\(\Leftrightarrow3x-1\in\left\{5;-5\right\}\)

hay \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

c: \(\Leftrightarrow\left(2x-5\right)^3=-81\)

\(\Leftrightarrow2x-5=-3\sqrt[3]{3}\)

hay \(x=\dfrac{5-\sqrt[3]{3}}{2}\)

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12

Bài 1 : 

a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18-2\right)\)

\(=6x^2+19x-7-6x^2-x+5-16=18x-18\)

Vậy biểu thức phụ thuộc biến x 

b, \(\left(x-2\right)\left(x+1\right)\left(2x+1\right)-x\left(2x^2-x-5\right)+1\)

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

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

Vậy biểu thức phụ thuộc biến x 

a)\(2x\left(x-2016\right)-2x+4032=0\)

\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)

\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)

b)\(5x\left(x-3\right)=x-3\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)

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

\(\Leftrightarrow\left(3x-1\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)

1. (-2x - 1)(x² - x - 3) - (x + 2)(x + 1)²

= -2x³ + 2x² + 6x - x² + x + 3 - (x + 2)(x² + 2x + 1)

= -2x³ + x² + 7x + 3 - x³ - 2x² - x - 2x² - 2x - 2

= -3x³ - 3x² + 4x + 1

2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3

=> (x + 2)(x - 1 - x + 3) = 3

=> (x + 2).0 = 3

...(xem lại đề)

\(\left(x+2\right)\left(x-1\right)-\left(x-3\right)\left(x+2\right)=3\)

\(\Leftrightarrow\left(x+2\right)\left(x-1-x+3\right)=3\)

\(\Leftrightarrow2\left(x+2\right)=3\)

\(\Leftrightarrow x+2=\frac{3}{2}\)

\(\Leftrightarrow x=\frac{3}{2}-2\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(x^2+11x-13=0\)

Ta có: \(\Delta=11^2+4.13=173\)

Vậy pt có 2 nghiệm phân biệt:

\(x_1=\frac{-11+\sqrt{173}}{2}\);\(x_2=\frac{-11-\sqrt{173}}{2}\)

\(2x^2-5x+2=0\)

Ta có: \(\Delta=5^2-4.2.2=9,\sqrt{\Delta}=3\)

Vậy pt có 2 nghiệm phân biệt:

\(x_1=\frac{5+3}{4}=2\);\(x_2=\frac{5-3}{4}=\frac{1}{2}\)