Ta có:
\(x^3+\left(x-1\right)^3=\left(2x-1\right)^3\)
Áp dụng hằng đẳng thức, ta có:
\(\left(x+x-1\right)\left(x^2-x^2-x+x^2-2x+1\right)=\left(2x-1\right)\left(x^2-3x+1\right)=\left(2x-1\right)^3\)
\(x^2-3x+1=\left(2x-1\right)^2=4x^2-4x+1\)
\(3x^2-x=0\)
\(x=0\) hoặc \(x=\frac{1}{3}\)
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cách 2:
\(x^3+\left(x-1\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow\left[x+\left(x-1\right)\right]^3-3x\left(x-1\right)\left[x+\left(x-1\right)\right]=\left(2x-1\right)^3\)(a3 + b3 = (a + b)3 - 3ab(a+b))
\(\Leftrightarrow\left(2x-1\right)^3-3x\left(x-1\right)\left(2x-1\right)=\left(2x-1\right)^3\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\\x=\frac{1}{2}\end{array}\right.\)
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