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14 tháng 5 2019

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

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

b) khong hiểu đề

a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)

=>7/6x=2/3

hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)

hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

13 tháng 3 2020

\(1,\left|2x-3\right|=x-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\\left[{}\begin{matrix}2x-3=x-5\\2x-3=-x+5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}5\\\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)

=> pt vô nghiệm

\(2,\left|3x+2\right|=x+1\)

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

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

\(3,\left|2x+1\right|=7-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}7-x\text{≥}0\\\left[{}\begin{matrix}2x+1=7-x\\2x+1=x-7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}7\\\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\end{matrix}\right.\) (loại)

=> pt vô nghiệm

\(4,\left|2x-5\right|=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\end{matrix}\right.\)

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

\(5,\left|6x-2\right|=3x-4\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}\frac{4}{3}\\\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(6,\left|3x-2\right|=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\text{≥}0\\\left[{}\begin{matrix}3x-2=x-2\\3x-2=-x+2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}2\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(7,\left|2x+3\right|=1\)

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

\(8,\left|2-x\right|=2x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\left[{}\begin{matrix}2-x=2x-1\\2-x=-2x+1\end{matrix}\right.\end{matrix}\right.\)

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

\(9,\left|2x-1\right|=x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\\left[{}\begin{matrix}2x-1=x-3\\2x-1=-x+3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(10,2\left|x-1\right|=x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left[{}\begin{matrix}2x-2=x+2\\2x-2=-x-2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

5 tháng 12 2017

nik lộn

2x+1. 22009 = 22010

10 - 2x = 25 - 3x

15 tháng 4 2020

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

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