\(B=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3\left(x^2-4\right)}{2x^2-8x+8}\)

\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x^2-4x+4\right)}\)

\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{1}{3}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)

\(=\dfrac{x-2}{x+2}\cdot\dfrac{3\left(5x+10\right)+7\left(x-2\right)}{21\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)

\(=\dfrac{1}{x+2}\cdot\dfrac{15x+30+7x-14}{21}+\dfrac{3x+6}{2\left(x-2\right)}\)

\(=\dfrac{22x+16}{21\left(x+2\right)}+\dfrac{3x+6}{2\left(x-2\right)}\)

\(=\dfrac{2\left(x-2\right)\left(22x+16\right)+21\left(x+2\right)\left(3x+6\right)}{42\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(2x-4\right)\left(22x+16\right)+\left(21x+42\right)\left(3x+6\right)}{42\left(x^2-4\right)}\)

\(=\dfrac{44x^2+32x-88x-64+63x^2+126x+126x+252}{42x^2-168}\)

\(=\dfrac{107x^2+196x+188}{42x^2-168}\)

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

Giải:

a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)

\(\Leftrightarrow24x-16-13x=60-15x+7x\)

\(\Leftrightarrow24x-13x+15x-7x=60+16\)

\(\Leftrightarrow19x=76\)

\(\Leftrightarrow x=\dfrac{76}{19}=4\)

Vậy ...

b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)

ĐKXĐ: \(x\ne\pm2\)

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

\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)

\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)

\(\Leftrightarrow8x^2-13x=0\)

\(\Leftrightarrow x\left(8x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)

Vậy ...

c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)

ĐKXĐ: \(x\ne-1;x\ne3\)

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

\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)

\(\Leftrightarrow x\left(2x-2\right)=4x\)

\(\Leftrightarrow2x-2=4\)

\(\Leftrightarrow x=3\)

Vậy ...

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

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

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

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

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

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

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

\(=\dfrac{y\left(y-2x\right)}{3}\)

Câu 1:

Câu 2:

ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)

\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)

\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)

\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)

\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)

Vậy \(S=\left\{-1\right\}\)

a: =>10x-14=15-9x

=>19x=29

hay x=29/19

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x+9=32x+60

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>101x=101

hay x=1

d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)

\(\Leftrightarrow6-18x+5x-6=0\)

=>-13x=0

hay x=0

\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)

\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)

\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)

\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)

\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)