Question

Giải phương trình 

a)(x-1)^3+(x-2)^3+(3-2x)^3=0

b)x/x^2+4x+4+5x/x^2+4=-2

Answer 1

a, Ta có: \(\left(x-1\right)^3+\left(x-2\right)^3+\left(3-2x\right)^3=0\)

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

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

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

\(\Leftrightarrow3\left(x-1\right)\left(x-2\right)\left(2x-3\right)=0\)

\(x=1,2,\frac{3}{2}\)