`Answer:`

Ta có vế phải: `(a+b)^3 -3ab.(a+b)`

`=(a^3 +3a^2 b+3ab^2 +b^3)-(3a^2 b+3ab^2)`

`=a^3+3a^2 b+3ab^2 +b^3 -3a^2 b-3ab^2`

`=a^3 +b^3 +(3a^2 b-3a^2 b)+(3ab^2 -3ab^2)`

`=a^3+b^3`

`=` Vế trái

Vậy `a^3 +b^3 =(a+b)^3 -3ab.(a+b)`

`Answer:`

c) \(\frac{2\left(3x+5\right)}{3}-\frac{x}{x}=5-\frac{3\left(x+1\right)}{4}\)

\(\Leftrightarrow\frac{8\left(3x+5\right)}{12}-\frac{6x}{12}=\frac{60}{12}-\frac{9\left(x+1\right)}{12}\)

\(\Rightarrow24x+40-6x=60-9x-9\)

\(\Leftrightarrow24x+9x-6x=60-9-40\)

\(\Leftrightarrow27x=11\)

\(\Leftrightarrow x=\frac{11}{27}\)

d) \(x^2-4x+4\)

\(\Leftrightarrow x^2-2x-2x+4\)

\(\Leftrightarrow x\left(x-2\right)-2\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2\)

e) \(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow\frac{x-2}{x+2}-\frac{x}{x-2}=\frac{5x-8}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{5x-8}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)-x\left(x+2\right)=5x-8\)

\(\Leftrightarrow x^2-2x-x+2-x^2-2x=5x-8\)

\(\Leftrightarrow-5x+2=5x-8\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

\(a^4+b^4\ge ab^3+ba^3\)

\(\Leftrightarrow a^4+b^4-ab^3-ba^3\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(đúng)

Vậy ta có điều phải chứng minh.

`Answer:`

\(4x-\frac{2x-1}{4}\le\frac{3x-2}{6}\)

\(\Leftrightarrow\frac{16x-2x+1}{4}\le\frac{3x-2}{6}\)

\(\Leftrightarrow\frac{14x+1}{4}\le\frac{3x-2}{6}\)

\(\Leftrightarrow3.\left(14x+1\right)\le2.\left(3x-2\right)\)

\(\Leftrightarrow42x+3\le6x-4\)

\(\Leftrightarrow42x-6x\le-4-3\)

\(\Leftrightarrow36x\le-7\)

\(\Leftrightarrow x\le-\frac{7}{36}\)

\(4x\left(3x-2\right)=4-6x\)\(\Leftrightarrow4x\left(3x-2\right)+6x-4=0\)\(\Leftrightarrow4x\left(3x-2\right)+2\left(3x-2\right)=0\)\(\Leftrightarrow2\left(3x-2\right)\left(2x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{1}{2}\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{-\frac{1}{2};\frac{2}{3}\right\}\)