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

\(\Leftrightarrow9x-21=10x-5\)

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

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

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

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

<=> 1 - x + 3x + 3 = 2x + 3

<=> 1 - x + 3x + 3 - 2x = 3

<=> 4 = 3 (vô lý)

=> pt vô nghiệm

b) ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)

<=> 2x - x² - 4 + 2x - 5x - 5x² + 10 = 15x - 30

<=> -x + 4x² - 14 = 15x - 30

<=> x - 4x² + 14 = 15x - 30 

<=> x - 4x² + 14 + 15x - 30 = 0

<=> 16x - 4x² - 16 = 0

<=> 4(4x - x² - 4) = 0

<=> -x² + 4x - 4 = 0

<=> x² - 4x + 4 = 0

<=> (x - 2)² = 0

<=> x - 2 = 0

<=> x = 2 (ktm)

=> pt vô nghiệm 

c) xem bài 4 ở đây: Câu hỏi của gjfkm

d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)

\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)

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

<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)

<=> x² - 3x + 4x - 12 + x² - 2x + x - 2 = 2x² - 4x + 5x - 10

<=> 2x² - 14 = 2x² + x - 10

<=> 2x² - 14 - 2x² = x - 10

<=> -14 = x - 10

<=> -14 + 10 = x

<=> -4 = x

<=> x = -4

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)

\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)

\(< =>4x-2+x=14x+7\)

\(< =>14x-5x=-2-7\)

\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)

\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

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

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)

\(\frac{x-1}{4}-\frac{5-2x}{9}=3x-\frac{2}{3}\)

\(< =>\frac{\left(x-1\right).9}{36}-\frac{\left(5-2x\right).4}{36}=\frac{3x.36}{36}-\frac{2.12}{36}\)

\(< =>\left(x-1\right).9-\left(5-2x\right).4=108x-24\)

\(< =>9x-9-20+8x=108x-24\)

\(< =>108x-17x=-29+24\)

\(< =>91x=-5< =>x=-\frac{5}{91}\)

a, x( x - 1) = x ( x + 2)

<=> x² - x = x² + 2x

<=>  x² - x - x² - 2x = 0

<=> -3x = 0

<=> x = 0

b, tương tự câu a

c,\(\Leftrightarrow\frac{3x-3}{4}=2-\frac{x-2}{8}\)        

\(\Leftrightarrow\frac{\left(3x-3\right)2}{8}=\frac{16}{8}-\frac{x-2}{8}\)

\(\Leftrightarrow\frac{6x-6}{8}=\frac{16}{8}-\frac{x-2}{8}\)

=> 6x - 6 = 16 - x + 2

<=> 6x + x = 16 + 2 + 6

<=> 7x = 24

<=> x=\(\frac{24}{7}\)

Các câu còn lại làm tương tự

Bài 1:

d)ĐKXĐ: \(x\ne8\)

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

MTC=24(x-8)

\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)

\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)

\(\Leftrightarrow348-29x=0\)

\(\Leftrightarrow-29x+348=0\)

\(\Leftrightarrow x=\frac{-348}{-29}=12\)

Vậy: x=12

e) ĐKXĐ: \(x\ne\pm1\)

Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)

MTC=4(x+1)(x-1)

\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)

\(\Leftrightarrow20x+4=0\)

\(\Leftrightarrow20x=-4\)

\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)

Vậy: x không có giá trị

g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)

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

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

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

\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)

MTC=2(x+1)

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

\(\Leftrightarrow2-x+1=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)

Vậy: x không có giá trị