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29 tháng 12 2019

ĐKXĐ: \(\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)

\(\frac{\left|2x-1\right|}{\left(x+1\right)\left(x-2\right)}>\frac{1}{2}\) (*)

+) Nếu \(x>2\) thì (*) \(\Leftrightarrow\frac{2x-1}{x^2-x-2}>\frac{1}{2}\)

\(\Leftrightarrow4x-2>x^2-x-2\)

\(\Leftrightarrow x^2-5x< 0\)

\(\Leftrightarrow x\left(x-5\right)< 0\)

\(\Leftrightarrow0< x< 5\)

\(\Leftrightarrow2< x< 5\)

+) Nếu \(x< -1\) thì (*) \(\Leftrightarrow\frac{1-2x}{x^2-x-2}>\frac{1}{2}\)

\(\Leftrightarrow2-4x>x^2-x-2\)

\(\Leftrightarrow x^2+3x-4< 0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)< 0\)

\(\Leftrightarrow-4< x< 1\)

\(\Leftrightarrow-4< x< -1\)

Vậy...

NV
26 tháng 10 2019

a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)

b/

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)

\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)

\(\Leftrightarrow\left|x+2\right|-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

c/

\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)

\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)

NV
27 tháng 10 2019

d/ ĐKXĐ: ...

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

Đặt \(\frac{\left|x-2\right|}{x-1}=a\)

\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)

e/ ĐKXĐ: ...

Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)

\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)

17 tháng 2 2017

Bai1:

\(-2x+\frac{3}{5}\le\frac{3\left(2x-7\right)}{3}\Leftrightarrow-10x+3\le5\left(2x-7\right)\Leftrightarrow-10x+3\le10x-35\)

\(\Leftrightarrow\left(10+10\right)x\ge3+35\Rightarrow x\ge\frac{38}{20}=\frac{19}{10}\)

Bài

\(\left\{\begin{matrix}x+m-1>0\\3m-2-x>0\end{matrix}\right.\Leftrightarrow\left(I\right)\left\{\begin{matrix}x>1-m\\x< 3m-2\end{matrix}\right.\)

Hệ (I) có nghiệm cần m thỏa mãn:

\(1-m< 3m-2\Leftrightarrow1+2< 3m+m\Rightarrow m>\frac{3}{2}\)

Kết luận: để hệ có nghiệm cần: m>3/2

23 tháng 3 2020
https://i.imgur.com/SmYpZ8d.jpg
23 tháng 3 2020
https://i.imgur.com/D95iizc.jpg