Question

[Lê Ngọc Minh Châu]

(x²-1)³+(x²+2)³+(2x-1)³+(3x+3)(2x-1)(1-x)(x²+2)=0

Answer 1

[Phạm Thành Đông]

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

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

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

Đặt \(x^2-1=a\), \(x^2+2=b\), \(2x-1=c\). Phương trình trở thành:

\(a^3+b^3+c^3-3abc=0\).

\(\Leftrightarrow\left[a^3+b^3+3ab\left(a+b\right)\right]-3ab\left(a+b\right)+c^3-3abc=0\).

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\).

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\).

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\).

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\).

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\).

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\).

Xét phương trình \(\left(1\right)\).

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

\(\Leftrightarrow2x^2+2x=0\).

\(\Leftrightarrow2x\left(x+1\right)=0\).

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\).

Xét phương trình \(\left(2\right)\)

\(\left(2\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0.2\).

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\).

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\).

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

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

\(\Leftrightarrow\left[\left(x-1\right)^2+2\right]^2+\left(x^2-2x\right)^2+9=0\).

Ta có:

\(\left(x-1\right)^2\ge0\forall x\).

\(\Rightarrow\left(x-1\right)^2+2\ge2\forall x\).

\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2\ge4\forall x\).

\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9\ge13\forall x\).

\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9>0\forall x\).

\(\left(x^2-2x\right)^2\ge0\forall x\).

\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9+\left(x^2-2x\right)^2>0\forall x\).

Do đó phương trình \(\left(2\right)\)vô nghiệm.

Vậy phương trình có tập nghiệm: \(S=\left\{0;-1\right\}\).

\(\)