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NV
11 tháng 2 2020

\(\frac{x^2+3x-1}{2-x}+x>0\Leftrightarrow\frac{5x-1}{2-x}>0\Rightarrow\frac{1}{5}< x< 2\)

\(\frac{\left(x-1\right)^3\left(x+2\right)^2\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x=-2\\1\le x< 2\\2< x< 7\end{matrix}\right.\)

Kết hợp lại ta có: \(1\le x< 2\)

NV
17 tháng 9 2019

a/ ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-9\)

\(\Leftrightarrow2x=6\Rightarrow x=3\) (ktm)

Vậy pt vô nghiệm

b/ ĐKXĐ: \(x\ne1\)

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

\(\Leftrightarrow x^2+x+1+2\left(x-1\right)=3x^2\)

\(\Leftrightarrow2x^2-3x+1=0\Rightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=\frac{1}{2}\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ne\pm4\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{\left(x-4\right)\left(x+4\right)}+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow5x^2-80+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2+16=5x^2+2x\)

\(\Rightarrow x=8\)

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

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

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

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

giải các hệ BPT sau: a) \(\left\{{}\begin{matrix}5x-24x+5\\5x-4< x+2\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\) e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\) f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\) g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\) h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\) j)...
Đọc tiếp

giải các hệ BPT sau:

a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)

f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)

j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)

3
25 tháng 3 2020
https://i.imgur.com/NOxfqjV.jpg
25 tháng 3 2020
https://i.imgur.com/awOKwJi.jpg