Question

[Nguyễn Thùy Dương]

giải pt : \(\left(8x-11\right)^3+\left(7x-12\right)^2+\left(23-15x\right)^3=0\)

Answer 1

[Lê Thị Thục Hiền]

Sửa đề \(\left(8x-11\right)^3+\left(7x-12\right)^3+\left(23-15x\right)^3=0\)

Đặt \(8x-11=a\)

\(7x-12=b\)

\(23-15x=c\)

=> a+b+c=8x-11+7x-12+23-15x=0

Có \(a^3+b^3+c^3-3abc\)

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

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

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

=0 (do a+b+c=0)

=> \(a^3+b^3+c^3=3abc\)

<=> \(0=3\left(8x-11\right)\left(7x-12\right)\left(23-15x\right)\)

=> \(\left[{}\begin{matrix}x=\frac{11}{8}\\x=\frac{12}{7}\\x=\frac{23}{15}\end{matrix}\right.\)