`Answer:`

Bài 1:

a) \(7+2x=22-3x\)

\(\Leftrightarrow2x+3x=22-7\)

\(\Leftrightarrow5x=15\)

\(\Leftrightarrow x=3\)

b) \(8x-3=5x+12\)

\(\Leftrightarrow8x-5x=12+3\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

c) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow x-12+4x-25-2x+1=0\)

\(\Leftrightarrow\left(x+4x-2x\right)+\left(1-12-25\right)=0\)

\(\Leftrightarrow3x-36=0\)

\(\Leftrightarrow x=12\)

d) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow6x-19=3x+5\)

\(\Leftrightarrow6x-3x=5+19\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

Bài 2:

a) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}}\)

b) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow2x+7=0\text{ hoặc }x-5=0\text{ hoặc }5x+1=0\)

\(\Leftrightarrow x=-\frac{7}{2}\text{ hoặc }x=5\text{ hoặc }x=-\frac{1}{5}\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x^2=-1\text{(Loại)}\end{cases}}}\)

d) \(\left(x^2-4\right)+\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow x^2-4+\left(3x-2x^2-6+4x\right)=0\)

\(\Leftrightarrow x^2-4=\left(-2x^2+7x-6\right)=0\)

\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)

\(\Leftrightarrow-x^2+7x-10=0\)

\(\Leftrightarrow x^2-5x-2x+10=0\)

\(\Leftrightarrow x.\left(x-5\right)-2.\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right).\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

`Answer:`

`(3x+1)^2 -(x+1)`

`=(3x)^2 +2.3x.1+1^2 -x-1`

`=9x^2 +6x+1-x-1`

`=9x^2 +(6x-x)+(1-1)`

`=9x^2 +5x`

`5x^2+5xy-x-y`

`=(5x^2+5xy)-(x+y)`

`=5x.(x+y)-(x+y)`

`=(5x-1)(x+y)`

`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)`