a) (3x + 2)² - (3x - 2)² = 5x + 38

<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2

b) 3(x - 2)² + 9(x - 1) = 3(x² + x - 3)

<=> 3x² - 12x + 12 + 9x - 9 = 3x² + 3x - 9

<=> -6x = -12 <=> x = 2

c) (x + 3)² - (x - 3)² = 6x + 8

<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3

d) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> 3x = -21 <=> x = -7

e) (x + 1)(x² - x + 1) - 2x = x(x - 1)(x + 1)

<=> x³ - 1 - 2x = x³ - x

<=> x = -1

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)

\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

\(\Leftrightarrow\left[\left(3x+2\right)-\left(3x-2\right)\right]\left[\left(3x+2\right)+\left(3x-2\right)\right]=5x+38\)

\(\Leftrightarrow\left(3x+2-3x+2\right)\left(3x+2+3x-2\right)=5x+38\)

\(\Leftrightarrow4\cdot6x=5x+38\)

\(\Leftrightarrow24x-5x=38\)

\(\Leftrightarrow19x=38\Leftrightarrow x=\dfrac{38}{19}=2\)

Vậy \(S=\left\{2\right\}\)

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

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

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

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

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

PT vô nghiệm , không tìm được x 

Vậy \(S=\varnothing\)

c) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-2x+4\right)+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3x^2-6x+12+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-6x+12+9x-9-3x^2-3x+9=0\)

\(\Leftrightarrow0x+12=0\)

PT vô nghiệm 

Vậy \(S=\varnothing\)

Câu cuối tương tự 

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

<=> x³ - 6x² + 12x - 8 + 9x² - 1 - x³ - 3x² - 3x - 1 = 0

<=> 9x - 10 = 0

<=> 9x = 10

<=> x = 10/9

Vậy S = {10/9}

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

<=> 2x² - x - 3 - 2x² - 9x + 5 = 0

<=> -10x + 2 = 0

<=> -10x = -2

<=> x = 1/5

Vậy S = {1/5}

c) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> -5x² + 2x + 5x² + x + 22 - 1 = 0

<=> 3x = -21

<=> x = -7

Vậy S = {-7}

d) (x - 3)(x + 4) - 2(3x - 2) = (x - 4)²

<=> x² + x - 12 - 6x + 4 - x² + 8x - 16 = 0

<=> 3x - 24 = 0

<=> 3x = 24

<=> x = 8

Vậy S = {8}

e) x(x + 3)² - 3x = (x + 2)³ + 1

<=> x³ + 6x² + 9x - 3x = x³ + 6x² + 12x + 8 + 1

<=> x³ + 6x² + 6x - x³ - 6x² - 12x = 9

<=> -6x = 9

<=> x = -3/2

Vậy S = {-3/2}

f) (x + 1)(x² - x + 1) - 2x = x(x + 1)(x- 1)

<=> x³ + 1 - 2x = x³ - x

<=> x³ - 2x - x³ + x = -1

<=> -x = -1

<=> x = 1

Vậy S = {1}

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

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 \(x\in R\)

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