Thấy x = 0 không phải là n₀ của pt

=> pt <=> \(\dfrac{4}{x-8+\dfrac{7}{x}}\) +\(\dfrac{5}{x-10+\dfrac{7}{x}}\) = -1

Đặt x - 9 + \(\dfrac{7}{x}\) = a

=> pt <=> \(\dfrac{4}{a+1}\) + \(\dfrac{5}{a-1}\) = -1

<=> \(\dfrac{9a+1}{\left(a-1\right)\left(a+1\right)}\) = -1

<=> \(\dfrac{9a+1}{a^2-1}\) = -1

<=> 9a + 1 = 1 - a²

<=> a² + 9a = 0

<=> a(a + 9) = 0

TH1 a = 0 => x - 9 + \(\dfrac{7}{x}\) = 0

<=> x² - 9x + 7 = 0

<=> ( x - \(\dfrac{9}{2}\) )² = \(\dfrac{53}{4}\)

<=> x = \(\dfrac{9\pm\sqrt{53}}{2}\)

TH2 a = -9 => x - 9 + \(\dfrac{7}{x}\) = -9

<=> x² - 9x + 7 = -9x

<=> x² + 7 = 0 (vô lý)

Vậy x = \(\dfrac{9\pm\sqrt{53}}{2}\)

\(\Leftrightarrow\dfrac{2}{-x^2+6x-8}=\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\\ \Leftrightarrow\left\{{}\begin{matrix}2=\left(-x^2+6x-8\right)\left(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\right)\\-x^2+6x-8\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2=-2x^2+4x+2\\-x^2+6x-8\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\-x^2+6x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\-x^2+6x-8\ne0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\-x^2+6x-8\ne\end{matrix}\right.\end{matrix}\right.\\\Rightarrow x=0\)

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)

bỏ số 1 ở đầu thì giải dc á, còn có số 1 thì chịu

\(\dfrac{1}{x+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=\dfrac{4}{105}\)

\(\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{8}{x.\left(x+8\right)}=\dfrac{8}{105}\)

\(\Rightarrow x\left(x+8\right)=105\)

\(x^2+8x-105=0\)

\(\left(x-7\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

không có số 1 ở đầu đâu.Mong mọi người giải giúp mk nhé!

a: \(\Leftrightarrow x-5-2\left(2x-1\right)< 12\)

=>x-5-4x+2<12

=>-3x-3<12

=>-3x<15

hay x>-5

b: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)>10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12+15x-5-20x^2-30x+100>0\)

=>-15x+83>0

hay x<83/15

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)