PN
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
BB
1
8 tháng 5 2018
\(|\frac{1}{2}x+1|-4=0\)
\(\Rightarrow|\frac{1}{2}x+1|=4\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+1=4\\\frac{1}{2}x+1=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)
Vậy x = 6 hoặc x = -10
_Chúc bạn học tốt_
M
23 tháng 11 2018
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
22 tháng 11 2018
Tại vì nó được đề bài cho nên có nghĩa,k có nghĩa thì lm kiểu đếch j?
BL
10 tháng 3 2020
1. \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)
\(\Rightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\)
\(\Rightarrow x^2-x-2x+2+2x^2-2x-5x+5-3x^2+6x+5x-10=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
2. \(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}-\frac{16}{5}=0\)
\(\Rightarrow x^2-4x-3x+12-x^2+4x-4-16=0\)
\(\Rightarrow-3x-8=0\Rightarrow x=\frac{-8}{3}\)
3. \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{x^2-4}-\frac{3\left(x+2\right)}{x^2-4}-\frac{2\left(x-11\right)}{x^2-4}=0\)
\(\Rightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Rightarrow x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
6 tháng 7 2017
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)
19 tháng 3 2020
a, \(\frac{5x-2}{6}+\frac{3-4x}{2}=2\)\(-\frac{x+7}{3}\)
<=> \(\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)
=> 5x-2+9-12x = 12-2x-14
<=> 5x-12x+2x = 12-14-9+2
<=> -5x = -9
<=> x = \(\frac{9}{5}\)
\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) \(\left(x\ne1;x\ne-2\right)\)
\(\Rightarrow\frac{2\left(x-1\right)+2}{x-1}=\frac{\left(x+2\right)+2x}{x+2}\)\(\Rightarrow2x^2+4x=3x^2+2x-3x+2\)
\(\Rightarrow\frac{2x-2+2}{x-1}=\frac{x+2+2x}{x+2}\)
\(\Rightarrow\frac{2x}{x-1}=\frac{3x+2}{x+2}\)
\(\Rightarrow2x\left(x+2\right)=\left(x-1\right)\left(3x+2\right)\)
\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)
\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)
\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-2\end{cases}}\)
=> \(\frac{2\left(x-1\right)+2}{x-1}=\frac{x+2+2x}{x+2}\)
=> \(\frac{2\left(x-1+1\right)}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)
=> \(\frac{2x}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)
=> \(2x\left(x+2\right)=x+2\left(x+1\right)\left(x-1\right)\)
=> \(2x^2+4x=x+2\left(x^2-1\right)\)
=> \(2x^2+4x=x+2x^2-2\)
=> \(2x^2+4x-x-2x^2+2=0\)
=> \(3x+2=0\)
=> \(3x=-2\)
=> \(x=-\frac{2}{3}\)