a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)

=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)

=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)

=> \(6x+6+3x-6=12-8x+8\)

=> \(17x=20\)

=> \(x=\frac{20}{17}\)

b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)

=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)

=> \(4\left(11x-1\right)=6\left(6-x\right)\)

=> \(44x-4-36+6x=0\)

=> \(\)\(50x=40\)

=> \(x=\frac{4}{5}\)

c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)

=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)

=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)

=> \(20-40x+6x-9x+45+24=0\)

=> \(43x=89\)

=> \(x=\frac{89}{43}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)

\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)

Tập nghiệm: \(S=\left\{0\right\}\)

Bài 1 :

\(\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\left(ĐKXĐ:x\ne3\right)\)

\(\Leftrightarrow5\left(x^3-9x\right)=-\left(x^2+3x\right)\left(15-5x\right)\)

\(\Leftrightarrow5x^3-45x=5x^3-45\) ( luôn đúng )

Do đó : \(\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\left(x\ne3\right)\)

P/s : Bài này thì xét tích chéo của hai số thôi nhé @

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

a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)

a/b= 7/6 + 7/10 + 7/12

a/b= 7(1/6+1/10+1/12)

Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)

Bạn ơi cho mình hỏi dpcm là gì vậy?

dpcm: điều phải chứng minh

c) \(\frac{1}{2}-x=\frac{3}{4}\)

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

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

\(\Leftrightarrow x=\frac{-1}{4}\)

b) \(\frac{4}{5}x=\frac{4}{7}\)

\(\Leftrightarrow x=\frac{4}{7}\div\frac{4}{5}\)

\(\Leftrightarrow x=\frac{4}{7}.\frac{5}{4}\)

\(\Leftrightarrow x=\frac{5}{7}\)

a) \(\frac{3}{7}\div\frac{1}{4}+\frac{6}{11}x=\frac{4}{9}\)

\(\Leftrightarrow\frac{3}{7}\times4+\frac{6}{11}x=\frac{4}{9}\)

\(\Leftrightarrow\frac{12}{7}+\frac{6}{11}x=\frac{4}{9}\)

\(\Leftrightarrow\frac{6}{11}x=\frac{4}{9}-\frac{12}{7}\)

\(\Leftrightarrow\frac{6}{11}x=\frac{28}{63}-\frac{108}{63}\)

\(\Leftrightarrow\frac{6}{11}x=\frac{-80}{63}\)

\(\Leftrightarrow x=\frac{-80}{63}\div\frac{6}{11}\)

\(\Leftrightarrow x=\frac{-80}{63}.\frac{11}{6}\)

\(\Leftrightarrow x=\frac{-880}{378}=\frac{-440}{189}\)

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

h.3x - 2/6 - 5 = 3 - 2(x + 7)/4

<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4

<=> 3x - 32/6 = 3 - 2x - 14/4

<=> 3x - 32/6 = -2x - 11/4

<=> 6x - 64/12 = -6x - 33/12

<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12