giải pt sau
/x+1/ + /x-1/=4
K
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22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
NT
1
KN
13 tháng 2 2020
\(\left(2-x\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=-4\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)=-4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=-4\)
Đặt \(x^2-4=u\)
Phương trình trở thành \(u\left(u+3\right)=-4\)
\(\Leftrightarrow u^2+3u+4=0\)
Mà \(u^2+3u+4=\left(i^2+3u+\frac{9}{4}\right)+\frac{7}{4}=\left(u+\frac{3}{2}\right)^2+\frac{7}{4}>0\)nên phương trình vô nghiệm
NT
1
KN
14 tháng 2 2020
Consider two cases:
+) If \(x\ge2\)then \(x-2\ge0\Rightarrow\left|x-2\right|=x-2\)
Equation becomes: \(\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=4\)(1)
Put \(x^2-4=u\)
(1) becomes: \(u\left(u+3\right)=4\)
\(\Leftrightarrow u^2+3u-4=0\)
We have \(\Delta=3^2+4.4=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}u=\frac{-3+5}{2}=1\\u=\frac{-3-5}{2}=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=5\\x^2=0\end{cases}}\Rightarrow x\in\left\{\pm\sqrt{5};0\right\}\)
+) If \(x< 2\)then \(x-2< 0\Rightarrow\left|x-2\right|=2-x\)
Equation becomes: \(\left(2-x\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=-4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=-4\)(2)
Put \(x^2-4=v\)
(2) becomes: \(v\left(v+3\right)=-4\)
\(\Leftrightarrow v^2+3v+4=0\)
But \(v^2+3v+4=\left(v+\frac{3}{2}\right)^2+\frac{7}{4}>0\)
So case two has no value
So \(x\in\left\{\pm\sqrt{5};0\right\}\)
2 tháng 5 2017
\(\dfrac{x+4}{x+1}+\dfrac{x}{x-1}=\dfrac{2x^2}{x^2-1}\) ĐKXĐ: \(x\ne1;x\ne-1\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x^2+3x-4+x^2+1=2x^2\)
\(\Leftrightarrow x^2+x^2-2x^2+3x=4-1\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
22 tháng 4 2022
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
NV
17 tháng 5 2016
\(\frac{3}{x-1}+\frac{4}{x+1}=\frac{3x+2}{1-x^2}\Leftrightarrow\frac{3}{x-1}+\frac{4}{x+1}=-\frac{3x+2}{x^2-1}\Leftrightarrow\frac{3}{x-1}+\frac{4}{x+1}=-\frac{3x+2}{\left(x-1\right)\left(x+1\right)}\)\(\Leftrightarrow3.\left(x+1\right)+4.\left(x-1\right)=-\left(3x+2\right)\)
=> 3x + 3 + 4x - 4 + 3x + 2 = 0
=> 10x + 1 = 0
=> x = -1/10
LP
1
DT
Đỗ Thanh Hải
CTVVIP
10 tháng 3 2021
Ta có:
\(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\) (đk \(x\ge0,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\8\sqrt{x}-2\sqrt{y-1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}=9\\\sqrt{x}+2\sqrt{y-1}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\1+2\sqrt{y-1}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2\sqrt{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\left(tm\right)\)
HN
2
20 tháng 5 2022
\(\Leftrightarrow x-1-4=5\left(x-5\right)\)
=>x-5=5(x-5)
=>x-5-5x+25=0
=>-4x+20=0
hay x=5(loại)
+) Xét \(x\ge1\) có:
\(x+1+x-1=4\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( t/m )
+) Xét \(-1\le x< 1\) ta có:
\(x+1+1-x=4\)
\(\Leftrightarrow2=4\) ( vô lí )
+) Xét \(x< -1\) có:
\(-x-1+1-x=4\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\) ( t/m )
Vậy x = 2 hoặc x = -2
xét trường hợp : TH1: \(x\ge1\)
\(\Rightarrow\) \(\left|x+1\right|+\left|x-1\right|\) \(\Leftrightarrow\) \(x+1+x-1=4\) \(\Leftrightarrow\) \(2x=4\) \(\Leftrightarrow\) \(x=2\)
TH2 : \(x\le-1\)
\(\Rightarrow\) \(\left|x+1\right|+\left|x-1\right|\) \(\Leftrightarrow\) \(-x-1+1-x=4\) \(\Leftrightarrow\) \(-2x=4\) \(\Leftrightarrow\) \(x=-2\)vậy : \(x=-2;x=2\)