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/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

https://hoc24.vn/hoi-dap/question/655171.html

Lần sau ghi cho rõ đề

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mk chua hoc bai nay

\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8=7\)

\(\Leftrightarrow x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

a) \(27x^3+27x^2+9x+1=64\)

\(\Rightarrow27x^3+27x^2+9x-63=0\)

\(\Rightarrow27x^3-27x^2+54x^2-54x+63x-63=0\)

\(\Rightarrow27x^2\left(x-1\right)+54x\left(x-1\right)+63\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(27x^2+54x+63\right)=0\)

\(\Rightarrow\left(x-1\right).9\left(3x^2+6x+7\right)=0\)

\(\Rightarrow\left(x-1\right)\left(3x^2+6x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x^2+6x+7=0\end{matrix}\right.\)

Mà ta có:

\(3x^2+6x+7\)

\(=3\left(x^2+2x+\dfrac{7}{3}\right)\)

\(=3\left(x^2+2x+1-1+\dfrac{7}{3}\right)\)

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

Vì \(3\left(x+1\right)^2\ge0\) với mọi x

\(\Rightarrow3\left(x+1\right)^2+4\ge4\)

\(\Rightarrow3x^2+6x+7\) vô nghiệm

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

b) \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)

\(\Rightarrow12x-8=4\)

\(\Rightarrow12x=12\)

\(\Rightarrow x=1\)

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

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

\(\Rightarrow x^3-3x^2+3x-1-x^3-9+3x^2-12=2\)

\(\Rightarrow3x-22=2\)

\(\Rightarrow3x=24\)

\(\Rightarrow x=8\)