Ta có: \(\left(x-2\right)^3+\left(5-2x\right)^3=0\)

\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)

\(\Leftrightarrow3-x=0\)

hay x=3

e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

e)x³-4x+14x(x-2)=0

⇔ x(x²-4)+14x(x-2)=0

⇔ x(x-2)(x+2)+14x(x-2)=0

⇔ (x-2)(x²+2x+14x)=0

⇔ x(x-2)(x+16)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)

g)x²(x+1)-x(x+1)+x(x-1)=0

⇔ (x+1)(x²-x)+x(x-1)=0

⇔ x(x+1)(x-1)+x(x-1)=0

⇔ x(x-1)(x+2)=0

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

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

Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)

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

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

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

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

Ta có: \(2x^3+3x^2+2x+3=0\)

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

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

a:Ta có: \(2\left(x-2\right)^3=2-x\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

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

Suy ra: x-2=0

hay x=2

a: \(M=m^2\left(m+n\right)-n^2m-n^3\)

\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)

\(=\left(m+n\right)^2\left(m-n\right)\)

\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)

=0

b: \(N=n^3-3n^2-n\left(3-n\right)\)

\(=n^2\left(n-3\right)+n\left(n-3\right)\)

\(=n\left(n-3\right)\left(n+1\right)\)

\(=13\cdot10\cdot14=1820\)

a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)

(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)

(do (x+3)²+4≥4>0)

a: Ta có: \(x^3+3x^2+3x=0\)

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

hay x=0

b: Ta có: \(x^3+6x^2+12x=0\)

\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)

hay x=0

a: Ta có: \(2-x=2\left(x-2\right)^3\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)

bài a) bn trên đã dẫn link cho bn r

bài b)

Đặt x-y=a;y-z=b;z-x=c 

\(=>a+b+c=x-y+y-z+z-x=0\)

\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)

Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)

\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

a) Ta có :

\(a^3+b^3+c^3-3abc\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

P/s tham khảo nha

hok tốt

\(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)