a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

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

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0

=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0

=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0

=> -24x + 7 = 0 

=> - 24x = -7

=> x = 7/24

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

=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5

=> 6x - 5 = -5

=> 6x = 0

=> x = 0

c, x^2 = -6x - 8

=> x^2 + 6x + 8 = 0

=> x^2 + 2.x.3 + 9 - 1 = 0

=> (x + 3)^2 = 1

=> x + 3 = 1 hoặc x + 3 = -1

=> x = -2 hoặc x = -4

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

len google di ban

mk chua hoc bai nay

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

\(3x=5\)

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

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt