a) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

làm nốt đi 

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

a/ \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\frac{3}{2}\)

b/\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=19-22-28+15\)

\(\Leftrightarrow-8x=-16\)

\(\Leftrightarrow x=2\)

c/ \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7\left(2x-1\right)-3\left(5x+2\right)-21\left(x+13\right)}{21}=0\)

\(\Leftrightarrow14x-7-15x-6-21x-273=0\)

\(\Leftrightarrow-22x-286=0\)

\(\Leftrightarrow x=-13\)

e/ \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

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

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

\(\Leftrightarrow\frac{2\left(x^2-4\right)-\left(x^2+3x+2\right)-\left(3x^2-17x+22\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow2x^2-8-x^2-3x-2-3x^2+17x-22=0\)

\(\Leftrightarrow-2x^2+14x-32=0\)

\(\Leftrightarrow x^2-7x+16=0\)

\(\Leftrightarrow x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4.1.16}}{2}\)

\(\Leftrightarrow x=\frac{7\pm\sqrt{-15}}{2}\left(ktm\right)\)

\(\Leftrightarrow x\in\varnothing\)

Bài 1:

a) \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=18:12\)

\(\Leftrightarrow x=\frac{3}{2}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{3}{2}\right\}.\)

b) \(5.\left(2x-3\right)-4.\left(5x-7\right)=19-2.\left(x+11\right)\)

\(\Leftrightarrow10x-15-\left(20x-28\right)=19-\left(2x+22\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow13-10x=-3-2x\)

\(\Leftrightarrow13+3=-2x+10x\)

\(\Leftrightarrow16=8x\)

\(\Leftrightarrow x=16:8\)

\(\Leftrightarrow x=2.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)

c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7.\left(2x-1\right)}{7.3}-\frac{3.\left(5x+2\right)}{3.7}=\frac{21.\left(x+13\right)}{21}\)

\(\Leftrightarrow\frac{14x-7}{21}-\frac{15x+6}{21}=\frac{21x+273}{21}\)

\(\Leftrightarrow14x-7-\left(15x+6\right)=21x+273\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow-x-13=21x+273\)

\(\Leftrightarrow-x-21x=273+13\)

\(\Leftrightarrow-22x=286\)

\(\Leftrightarrow x=286:\left(-22\right)\)

\(\Leftrightarrow x=-13.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-13\right\}.\)

Chúc bạn học tốt!

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)

=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)

=> \(-\frac{3}{4}+\left(-2x\right)=-2\)

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

Vậy \(x\in\left\{\frac{5}{8}\right\}\)

\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)

=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)

=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)

=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)

=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)

Vậy \(x\in\left\{-\frac{39}{40}\right\}\)

\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)

=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)

( chiệt tiêu )

=> \(5x-6x+26=-14-7x\)

=> \(-x+26=-14-7x\)

=> \(-x+7x=-14-26\)

=> \(6x=-40\)

=> \(x=-40:6=\frac{20}{3}\)

Vậy \(x\in\left\{\frac{20}{3}\right\}\)

\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)

=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)

( chiệt tiêu )

=> \(2\left(2x-3\right)-9=5-3x-2\)

=> \(4x-6-9=3-3x\)

=> \(4x-15=3-3x\)

=> \(4x+3x=3+15\)

=> \(7x=18\)

=> \(x=18:7=\frac{18}{7}\)

Vậy \(x\in\left\{\frac{18}{7}\right\}\)

\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)

ĐKXĐ : \(x\ne0\)

=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)

=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)

=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)

=> \(\frac{32}{3x}=\frac{1}{4}\)

=> \(3x=32.4:1=128\)

=> \(x=128:3=\frac{128}{3}\)

Vậy \(x\in\left\{\frac{128}{3}\right\}\)

\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)

ĐKXĐ :\(x\ne1;\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)

=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)

=> \(\frac{26+5-2}{2\left(x-1\right)}\)

=> \(\frac{29}{2\left(x-1\right)}\)

\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)

=> \(x=\frac{19}{10}:2=\frac{19}{20}\)

Vậy \(x\in\left\{\frac{19}{20}\right\}\)

\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)

=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)

=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)

=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{1}{4}\right\}\)