\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

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

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

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

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)    ĐKXĐ : \(x\ne\pm2\)

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

\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2=2x^2+4\)

\(\Leftrightarrow0x=0\)

KL : PT vô số nghiệm 

AYUASGSHXHFSGDB HAGGAHAJF

A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)

A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)

Rồi tiếp tục làm nhé bạn.

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x² + 12x + 3 - 5x² - 10x - 5 = 7x² - 14x - 5

<=> 7x² + 2x - 7x² + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

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

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x² + 14x + 40 + 3x² + 6x - 24 = 4x² + 32x - 80

<=> 4x² + 20x - 4x² - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

\(\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^2+\left(\frac{x}{4}\right)^2=\left(\frac{x}{5}\right)^2+\left(\frac{x}{6}\right)^2+\left(\frac{x}{7}\right)^2\)

\(\frac{x^2}{2^2}+\frac{x^2}{3^2}+\frac{x^2}{4^2}=\frac{x^2}{5^2}+\frac{x^2}{6^2}+\frac{x^2}{7^2}\)

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

\(x^2.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}-\frac{1}{7^2}\right)=0\)

vì \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}-\frac{1}{7^2}\ne0\)nên \(x^2=0\)

\(\Rightarrow x=0\)

x^2(1/4+1/9+1/16-1/25-1/36/1/49)=0
mà (1/2+1/9=1/16-1/25-1/36-1/49)>0
=>x=0

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)

<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)

<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)

<=> \(3x^4+18x^2+12x-33=0\)

<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)

<=> \(x-1=0\)

<=> \(x=1\)

Mà vì: \(x^3+x^2+7x+11\ne0\)

=> x = 1

\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)

\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)

\(=>\left(x^2+3\right)^2=16\)

Đến đây tự tìm nha 

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