`Answer:`

Câu 1:

Câu 2:

a) \(\frac{2x+1}{6x-5}\ge\frac{3x-2}{9x-1}\)

\(\Leftrightarrow\left(2x+1\right)\left(9x-1\right)\ge\left(6x-5\right)\left(3x-2\right)\)

\(\Leftrightarrow18x-2x+9x-1\ge18x-12x-15x+10\)

\(\Leftrightarrow7x-1\ge-27x+10\)

\(\Leftrightarrow7x+27x\ge10+1\)

\(\Leftrightarrow-20x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{20}\)

b) \(\frac{3}{1-x}\le\frac{3}{2x+1}\left(x\ne1;x\ne-\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{3}{1-x}\ge0\)

\(\Leftrightarrow\frac{3\left(1-x\right)-3\left(2x+1\right)}{\left(2x+1\right)\left(1-x\right)}\ge0\)

Trường hợp 1: \(\hept{\begin{cases}3\left(1-x\right)-3\left(2x+1\right)\ge0\\\left(x+1\right)\left(1-x\right)>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 0\\-\frac{1}{2}< x< 1\end{cases}}\Leftrightarrow0< x< 1\)

Trường hợp 2: \(\hept{\begin{cases}3\left(1-x\right)-3\left(2x+1\right)< 0\\\left(2x+1\right)\left(1-x\right)< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x< -\frac{1}{2}\text{ hoặc }x>1\end{cases}}\Leftrightarrow x>1\)

Câu 3:

a) Để cho giá trị của biểu thức `\frac{2x+1}{x-2}` không lớn hơn `1`

\(\Leftrightarrow\frac{2x+1}{x-2}\le1\)

\(\Leftrightarrow2x+1\le x-2\)

\(\Leftrightarrow2x-x\le-2-1\)

\(\Leftrightarrow x\le-3\)

b) Để cho giá trị của biểu thức `\frac{3x+1}{2x-1}` không bé hơn `2`

\(\Leftrightarrow\frac{3x+1}{2x-1}\ge2\)

\(\Leftrightarrow3x+1\ge2\left(2x-1\right)\)

\(\Leftrightarrow3x+1\ge4x-2\)

\(\Leftrightarrow3x-4x\ge-2-1\)

\(\Leftrightarrow-x\ge-3\)

\(\Leftrightarrow x\le3\)

`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.