1/ 

a, \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b, \(B=2\dfrac{3}{11}.\dfrac{11}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{11}{12}.\left(-\dfrac{11}{5}\right)=-\dfrac{55}{12}\)

c, \(C=\left(\dfrac{3}{4}-0,2\right):\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right):\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}:\left(-\dfrac{2}{5}\right)=-\dfrac{11}{8}\)

2/ 

a, \(\dfrac{11}{12}-x=\dfrac{2}{3}+\dfrac{1}{4}\\ \Rightarrow\dfrac{11}{12}-x=\dfrac{11}{12}\\ \Rightarrow x=0\)

b, \(2x\left(x-\dfrac{1}{7}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\\ \Rightarrow x=-\dfrac{5}{7}\)

`a)-5/6-x=7/12+[-1]/3`

   `-5/6-x=1/4`

   `x=-5/6-1/4`

   `x=-13/12`

`b)(4,5-2x).(-1 4/7)=11/14`

   `(4,5-2x).(-11/7)=11/14`

    `4,5-2x=11/14:(-11/7)=-0,5`

    `2x=4,5-(-0,5)=5`

    `x=5:2=2,5`

`c)(2/11+1/3)x=(1/7-1/8).56`

   `17/33 x=1/56 .56=1`

      `x=33/17`

a.   2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0

2x=- 4/5 hoặc 3x=1/2

x=-2/5 hoặc x=\(\dfrac{1}{6}\)

b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0

x=2/5 hoặc x=-\(\dfrac{4}{7}\)

d. x(1+5/8-12/16)=1

\(\dfrac{7}{8}\)x=1=> x=8/7

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).

\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)

\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)

\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)

\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)

\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)

\(\Leftrightarrow x=1\left(koTM\right).\)

a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)

b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)

c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)

TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

Làm vs mn cần gấp

\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)

\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)

a, đk : x khác -1 ; x khác 2 

\(\Rightarrow4x-8=3x+3\Leftrightarrow x=11\left(tm\right)\)

b, đk : x khác 2 ; -2 

\(\Rightarrow\left(x+2\right)^2-8x=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\left(ktm\right)\)

-> vậy pt vô nghiệm 

c, đk : x khác 3  ; 0 

\(\Rightarrow x-5\left(x-3\right)=3x+7\Leftrightarrow-4x+15=3x+7\Leftrightarrow-7x=-8\Leftrightarrow x=\dfrac{8}{7}\left(tm\right)\)

a)Với \(x\ne-1;x\ne2\)

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

<=>4(x-2)=3(x+1)

<=>4x-8=3x+3

<=>x=11(TM)

b)Với\(x\ne\pm2\)

\(\dfrac{x+2}{2x-4}-\dfrac{4x}{x^2-4}=0\)

<=>\(\dfrac{x+2}{2\left(x-2\right)}-\dfrac{4x}{\left(x+2\right)\left(x-2\right)}=0\)

<=>\(\dfrac{\left(x+2\right)^2-8x}{2\left(x+2\right)\left(x-2\right)}=0\)

<=>\(x^2+4x+4-8x=0\left(Vĩx\ne\pm2\right)\)

<=>\(x^2-4x+4=0\)

<=>\(\left(x-2\right)^2=0\)

<=>x-2=0

<=>x=2(Không thỏa mãn)

c)Với \(x\ne3,x\ne0\)

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

<=>\(\dfrac{x-5\left(x-3\right)}{x\left(x-3\right)}=\dfrac{3x+7}{x\left(x-3\right)}\)

<=>x-5x+15=3x+7(Vì \(x\ne0,x\ne3\))

<=>7x=8

<=>x=\(\dfrac{8}{7}\left(TM\right)\)

\(a,\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\\ b,\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\\ c,\Rightarrow\dfrac{1}{4}:x=-\dfrac{2}{5}-\dfrac{3}{4}=-\dfrac{23}{20}\\ \Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{23}{20}\right)=-\dfrac{5}{23}\)