a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

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

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}

\(a,\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)\(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow5x-3=0\)

\(\Leftrightarrow x=\frac{3}{5}\left(tmđk\right)\)

Vậy ......

\(b,\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\) \(Đkxđ:....\)

\(\Leftrightarrow\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow-4x+8-16=10\)

\(\Leftrightarrow x=-\frac{9}{2}\)

Vậy ...............

ai giúp mk vs mai mk phải nôp bài

\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)

\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)

Vô nghiệm

a) ĐKXĐ: x∉{2;5}

Ta có: \(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

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

\(\Leftrightarrow6x+1+5x-25-3\left(x-2\right)=0\)

\(\Leftrightarrow11x-24-3x+6=0\)

\(\Leftrightarrow8x-18=0\)

\(\Leftrightarrow8x=18\)

hay \(x=\frac{9}{4}\)(tm)

Vậy: \(x=\frac{9}{4}\)

b) ĐKXĐ: x∉{0;2;-2}

Ta có: \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)

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

\(\Leftrightarrow2x-\left(x^2+x-2\right)+x^2-6x+8=0\)

\(\Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\)

\(\Leftrightarrow-5x+10=0\)

\(\Leftrightarrow-5x=-10\)

hay x=2(ktm)

Vậy: x∈∅

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

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

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) 

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

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé

Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà