\(\left|2x-1\right|< 3x+5\)

b)\(\left|x-1\...">

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4 tháng 3 2020

mình sửa lại bài 3 ý a, \(\left|5x-3\right|< 2\)

a: =>|x+3|=|2x-1|

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+3\\2x-1=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-2\end{matrix}\right.\Leftrightarrow x\in\left\{4;-\dfrac{2}{3}\right\}\)

b: \(\left|x^2-2x\right|=\left|2x^2-x-2\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+1\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1;-1;\dfrac{2}{3}\right\}\)

c: \(\left|3x^2-2x\right|=\left|6-x^2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\end{matrix}\right.\)

\(\Leftrightarrow2x^2-x-3=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\)

=>x=3/2 hoặc x=-1

d: \(\left|2x^2-3x-5\right|=\left|x^2-4x-5\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=x^2-4x-5\\2x^2-3x-5=4x+5-x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\\3x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\3x^2-10x+3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\\left(3x-10\right)\left(x+1\right)=0\end{matrix}\right.\)

hay \(x\in\left\{\dfrac{10}{3};-1\right\}\)

e: |5x+1|=|2x-3|

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=2x-3\\5x+1=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\7x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)

24 tháng 2 2020

giúp mình với mình đang cần gấp

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
NV
26 tháng 10 2019

a/

\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)

\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)

b/

\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)

\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)

c/

\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)

\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)

d/ \(x\ge\frac{3}{2}\)

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

\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)

NV
26 tháng 10 2019

e/

\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)

f/

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)

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

g/

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)