Ta có: \(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}x\)

<=> \(\frac{7}{8}-\frac{7}{10}=\frac{2}{3}x+\frac{1}{3}x\)

<=>\(\frac{7}{40}=x\)

Vậy x=7/40

\(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}\)

\(\Leftrightarrow-\frac{1}{3}x-\frac{2}{3}x=\frac{7}{10}-\frac{7}{8}\)

\(\Leftrightarrow-x=-\frac{7}{40}\)

\(\Leftrightarrow x=\frac{7}{40}=0,175\)

Ta có: \(\frac{7}{8}-\frac{1}{3}.x=\frac{7}{10}+\frac{2}{3}.x\)

\(\Rightarrow\frac{7}{8}-\frac{7}{10}=\frac{2}{3}.x+\frac{1}{3}.x\)

\(\Rightarrow\frac{35}{40}-\frac{28}{40}=\left(\frac{2}{3}+\frac{1}{3}\right).x\)

\(\Rightarrow\frac{7}{40}=x\)

Vậy \(x=\frac{7}{40}\)

Chúc bạn hok tốt! 

Khi đó phương trình đã cho tương đương với: \(4\left(\sqrt{x+2}-2\right)+\left(\sqrt{22-3x}-4\right)=x^2-4\)

\(\Leftrightarrow\frac{4\left(x-2\right)}{\sqrt{x+2}-2}+\frac{3\left(2-x\right)}{\sqrt{22-3x}+4}=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-\frac{4}{\sqrt{x+2}-2}+\frac{3}{\sqrt{22-3x}+4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2-\frac{4}{\sqrt{x+2}-2}+\frac{3}{\sqrt{22-3x}+4}=0\end{cases}\left(1\right)}\)

Xét hàm số f(x)=\(x+2-\frac{4}{\sqrt{x+2}-2}+\frac{3}{\sqrt{22-3x}+4}\left(-2\le x\le\frac{10}{3}\right)\)

Ta có \(f'\left(x\right)=1+\frac{2}{\sqrt{x+2}+\left(\sqrt{x+2}-2\right)}+\frac{9}{\sqrt{22-3x}\left(\sqrt{22-3x}+4\right)}>0\)với mọi \(x\in\left(-2;\frac{22}{3}\right)\)Do đó hàm f(x) đồng biến trên \(x\in\left[-2;\frac{22}{3}\right]\)

Mặt khác ta thấy f(-1)=0 nên x=-1 là nghiệm duy nhất của phương trình (1)

Vậy x=2;x=-1 là nghiệm của phương trình

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

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

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

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

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

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x-8}+\frac{1}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\end{cases}}\)

đến đoạn bạn giải tiếp nhé

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

cái đề hơi khó hiểu đó

theo đề ra ta có :

\(\left\{{}\begin{matrix}x=10+y\\\frac{160\left(x-y\right)}{xy}=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\frac{160.10}{\left(10+y\right)y}=0,8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10+y\\10y+y^2=\frac{1600}{0,8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\y^2+10y-2000=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\left(y-40\right)\left(y+50\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\left[{}\begin{matrix}y=40\\y=-50\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=40+10=50\\y=40\end{matrix}\right.\\\left\{{}\begin{matrix}x=-50+10=-40\\y=-50\end{matrix}\right.\end{matrix}\right.\)

Vậy .............

Từ đề bài, ta có:

\(2+\frac{x-30}{10}+2+\frac{x-28}{9}+2+\frac{x-26}{8}=0\)

\(\Leftrightarrow\frac{x-10}{10}+\frac{x-10}{9}+\frac{x-10}{8}=0\)

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

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

<=> x=10

Vậy phương trình có nghiệm duy nhất x=10

@Gà Mờ Thks bạn 

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

\(-537x^2+5054x=-541x^2+5092x\)

\(-537x^2+5054x+541x^2-5092x=0\)

\(4x^2-38x=0\)

\(x\left(2x-19\right)=0\)

\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)

ĐKXĐ: \(x\ne-2;-3;-4\)

Ta có: \(x+\frac{x}{x+2}+\frac{x+3}{x^2+5x+6}+\frac{x+4}{x^2+6x+8}=1\)

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

<=> \(\frac{x^2+2x}{x+2}+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}=1\)

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

<=>x=0

Vậy x=0

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1