`(2x)/(3x^2-x+2)-(7x)/(3x^2+5x+2)=1(x ne -1,-2/3)`

Đặt `a=3x^2+2x+2(a>=5/3)`

`pt<=>(2x)/(a-3x)-(7x)/(a+3x)=1`

`=>2x(a+3x)-7x(a-3x)=a^2-9x^2`

`<=>2ax+6x^2-7ax+21x^2=a^2-9x^2`

`<=>-5ax+27x^2=a^2-9x^2`

`<=>a^2-36x^2+5ax=0`

`<=>a^2-4ax+9ax-36x^2=0`

`<=>a(a-4x)+9x(a-4x)=0`

`<=>(a-4x)(a+9x)=0`

`+)a=4x`

`=>3x^2+2x+2=4x`

`=>3x^2-2x+2=0`

`=>x^2-2/3x+2/3=0`

`=>(x-1/3)^2+5/9=0` vô lý

`+)a+9x=0`

`=>3x^2+2x+2+9x=0`

`=>3x^2+11x+2=0`

`=>x^2+11/3x+2/3=0`

`=>x=(-11+-\sqrt{97})/6`

ĐKXĐ: \(x\ne-1;x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)(1)

\(\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{7}{3x+5+\dfrac{2}{x}}=1\)

Đặt: \(3x+\dfrac{2}{x}=a\)  (x khác 0) thì pt(1) trở thành:

\(\dfrac{2}{a-1}-\dfrac{7}{a+5}=1\)

\(\Leftrightarrow\dfrac{2\left(a+5\right)-7\left(a-1\right)}{\left(a-1\right)\left(a+5\right)}=1\)

\(\Leftrightarrow2\left(a+5\right)-7\left(a-1\right)=\left(a-1\right)\left(a+5\right)\)

\(\Leftrightarrow-5a+17=a^2+4a-5\)

\(\Leftrightarrow a^2+4a+5-5-17=0\)

\(\Leftrightarrow a^2+9a-22=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2}{x}=2\\3x+\dfrac{2}{x}=-11\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}3x^2+2-2x\ne0\\3x^2+11x+2\ne0\end{matrix}\right.\)

=> PT vô nghiệm 

Ủa hình như sai:vvv

a)

`2x-3=2-x`

`<=>2x+x=2+3`

`<=>3x=5`

`<=>x=5/3`

b)

`3x+3=7+5x`

`<=>3x-5x=7-3`

`<=>-2x=4`

`<=>x=-2`

c)

`7x-3=3x+13`

`<=>7x-3x=13+3`

`<=>4x=16`

`<=>x=4`

d)

`(5x-2)/3=(5-3x)/2`

`<=>10x-4=15-9x`

`<=>10x+9x=15+4`

`<=>19x=19`

`<=>x=1`

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

a, \(\Rightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)

b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

\(\Leftrightarrow21x+63-14=20x+36-49x+63\)

\(\Leftrightarrow50x=50\Leftrightarrow x=1\)

c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)

d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)

\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)

\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)

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):}`

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

Lời giải:

ĐKXĐ:.....

Ta có: \(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)

\(\Leftrightarrow \frac{1}{6}+\frac{2x}{3x^2-x+2}-7\left(\frac{x}{3x^2+5x+2}+\frac{1}{6}\right)=0\)

\(\Leftrightarrow \frac{3x^2+11x+2}{6(3x^2-x+2)}-\frac{7(3x^2+11x+2)}{6(3x^2+5x+2)}=0\)

\(\Leftrightarrow \frac{1}{6}(3x^2+11x+2)\left(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}\right)=0\)

TH1: \(3x^2+11x+2=0\)

\(\Leftrightarrow x=\frac{-11\pm \sqrt{97}}{6}\) (thỏa mãn)

TH2: \(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=0\)

\(\Leftrightarrow \frac{2}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=\frac{1}{3x^2-x+2}\)

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

\(\Leftrightarrow x=3x^2-x+2\)

\(\Leftrightarrow 3x^2-2x+2=0\)

\(\Leftrightarrow 2x^2+(x-1)^2+1=0\) (vô lý)

Do đó PT có nghiệm \(x=\frac{-11\pm \sqrt{97}}{6}\)

\(\dfrac{2x-1}{3x^2+7x+2}+\dfrac{3}{9x^2+15x+4}-\dfrac{2x+7}{3x^2-5x-12}=\dfrac{5}{x+2}\)

\(\Leftrightarrow\dfrac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\dfrac{3}{\left(3x+1\right)\left(3x+4\right)}-\dfrac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\dfrac{5}{x+2}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{3x+1}+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}+\dfrac{1}{3x+4}-\dfrac{1}{x-3}=\dfrac{5}{x+2}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x-3}=\dfrac{5}{x+2}\)

\(\Leftrightarrow\dfrac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow5x-3=-5\)

\(\Leftrightarrow x=-\dfrac{2}{5}\)

Vậy...