sao rối quá v bn ..

\(A=a^3-b^3-84\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-84\)

\(=\left(a-b\right)\left\{\left(a-b\right)^2+3ab\right\}\)

\(=6.\left[6^2+3.9\right]=6.63=379\)

\(Ủng\)hộ nhak

\(9x\left(2x-3\right)-6x+9=0\)

=>\(9x\left(2x-3\right)-3\left(2x-3\right)=0\)

=>\(\left(9x-3\right)\left(2x-3\right)=0\)

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

=>\(\orbr{\begin{cases}3x-1=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{3}{2}\end{cases}}}\)

Ta có:\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(y+z\right)+\left(y+z\right)^2\)

\(=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2\)

\(=x^2\)

\(=x.x\)

a) câu này dài quá à, mình ngại làm lắm

Áp dụng bđt này: \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

b)\(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)\)

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

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

\(=\left(1+2x^2+x^4-x^2\right)\left(1-x^2\right)\)

\(=\left(x^4+x^2+1\right)\left(1-x^2\right)\)

1/\(x^2+5x+6=0\)

=>\(x^2+2x+3x+6=0\)

=>\(x\left(x+2\right)+3\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)

Các câu sau làm tương tự câu 1, tách ghép khéo léo sẽ ra :)