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

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

\(5x^2-5x^2-5x=3\)

\(-5x=3\)

\(x=\frac{-3}{5}\)

tự ghi dấu suy ra ở đằng trước nhé

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

nên chỉ cần giải: \(6x^2-10x-3x+21=3\)

\(\Leftrightarrow6x^2-13x+21=3\)

\(\Leftrightarrow6x^2-13x+18=0\)

\(\Rightarrow\)pt vô nghiệm

a) (5x - 1)(2x + 1) = (5x -1)(x + 3)

<=> (5x - 1)(2x + 1) - (5x -1)(x + 3) = 0

<=> (5x - 1)(2x + 1 - x - 3) = 0

<=> (5x - 1)(x - 2) = 0

<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)

Vậy x = 0,2 ; x = 2 là nghiệm phương trình

b) x³ - 5x² - 3x + 15 = 0

<=> x²(x - 5) - 3(x - 5) = 0

<=> (x² - 3)(x - 5) = 0

<=> \(\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-5\right)=0\)

<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)

<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)

Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm

c) (x - 3)² - (5 - 2x)² = 0

<=> (x - 3 + 5 - 2x)(x - 3 - 5 + 2x) = 0

<=> (-x + 2)(3x - 8) = 0

<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)

Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)

d) x³ + 4x² + 4x = 0

<=> x(x² + 4x + 4) = 0

<=> x(x + 2)² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)

e) 3x² - 8x - 3 = 0

<=> 3x² - 9x + x - 3 = 0

<=> 3x(x - 3) + (x - 3) = 0

<=> (x - 3)(3x + 1) = 0

<=> \(\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{3;-\frac{1}{3}\right\}\)là giá trị cần tìm

f) (x² + 2x)² - 3(x² + 2x) - 4 = 0

<=> (x² + 2x)² - 3(x² + 2x) + 9/4 - 1/4 = 0

<=> \(\left(x^2+2x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=0\)

<=> \(\left(x^2+2x+\frac{3}{2}+\frac{1}{2}\right)\left(x^2+2x+\frac{3}{2}-\frac{1}{2}\right)=0\)

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

<=> (x + 1)² = 0 (Vì x² + 2x + 4 = (x + 1)² + 3 > 0 \(\forall\)x)

=> x + 1 = 0

<=> x = -1

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)

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

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

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

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

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)