a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)

\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)

\(\Leftrightarrow20x=-40\)

\(\Leftrightarrow x=-2\)

Vậy phương trình có nghiệm x = - 2

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)

\(\Rightarrow32x+32=31x+62\)

\(\Rightarrow x=30\)

Vậy x=30

Chúc bn học tốt

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

Ta có: \(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{1+2\cdot4}{2\cdot4}+\dfrac{1+3\cdot5}{3\cdot5}\cdot...\cdot\dfrac{1+x\left(x+2\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2\cdot2}{1\cdot3}+\dfrac{3\cdot3}{2\cdot4}+\dfrac{4\cdot4}{3\cdot5}+...+\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot2\cdot3\cdot...\cdot\left(x+1\right)}{1\cdot2\cdot3\cdot...\cdot x}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot\left(x+1\right)}{3\cdot4\cdot5\cdot...\cdot\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2x+2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{32x+32}{16\left(x+2\right)}=\dfrac{31\left(x+2\right)}{16\left(x+2\right)}\)

Suy ra: \(32x+32=31x+62\)

\(\Leftrightarrow x=30\)(thỏa ĐK)

Vậy: S={30}

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

`<=>(2x-1-2)(2x-1+2)=0`

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(x^2+x=t\)

\(\Rightarrow t\left(t+1\right)=42\)

\(\Leftrightarrow t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)

Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)

\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)

\(x+\frac{1}{2}=x+x+3\\\)

\(x+\frac{1}{2}=x+\left(x+3\right)\)

\(\Rightarrow\frac{1}{2}=x+3\)

\(\Rightarrow x=\frac{1}{2}-3\)

\(\Rightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)

\(Ta\) \(có\)

\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)

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

\(3x+2=4x\)

\(3x+2=3x+x\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) Vì \(\left|x+\frac{1}{5}\right|\ge0;\left|x+\frac{2}{5}\right|\ge0;\left|x+1\frac{2}{5}\right|\ge0\forall x\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

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

\(\Rightarrow3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)

\(\Rightarrow2=4x-3x=x\)

Vậy x = 2