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21 tháng 2 2021
a) Ta có: \(3-\left(17-x\right)=-12\)
\(\Leftrightarrow3-17+x+12=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)
\(\Leftrightarrow\left|x-9\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{24;-6\right\}\)
NB
21 tháng 11 2019
a)\(\left|4-2x\right|=6\\ \Rightarrow\left[{}\begin{matrix}4-2x=6\\2-2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
Vậy...
d)\(-12\left|x\right|=-24\\ \Leftrightarrow\left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy...
e)\(\left|x\right|\left(-3\right)=-9 \\ \Leftrightarrow\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy...
BV
0
NH
2
3 tháng 7 2019
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
=> \(\frac{2}{3}:x=-7-\frac{1}{3}\)
=> \(\frac{2}{3}:x=-\frac{22}{3}\)
=> \(x=\frac{2}{3}:\left(-\frac{22}{3}\right)\)
=> \(x=-\frac{1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
=> \(\frac{11}{15}x=0\)
=> \(x=0\)
c) \(\left(2x-3\right)\left(6-2x\right)=0\)
=> \(\left(2x-3\right)\left(3-x\right).2=0\)
=> \(\left(2x-3\right)\left(3-x\right)=0\)
=> \(\orbr{\begin{cases}2x-3=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
3 tháng 7 2019
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
\(\Rightarrow\frac{2}{3}.\frac{1}{x}=-7-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-21-1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-22}{3}\)
\(\Rightarrow-22.3x=6\)
\(\Rightarrow3x=\frac{-6}{22}=\frac{-3}{11}\)
\(\Rightarrow x=\frac{-3}{11}:3=\frac{-3}{11}.\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=0\)
\(\Rightarrow x=0\)
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
d) \(x:\frac{3}{4}+\frac{1}{4}=\frac{-2}{3}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-2}{3}-\frac{1}{4}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-11}{12}\)
\(\Rightarrow x=\frac{-11}{12}:\frac{4}{3}=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}\)
e) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)
16 tháng 6 2022
a: \(\Leftrightarrow2x-2+4x+8=-12\)
=>6x+6=-12
=>6x=-18
hay x=-3
b: \(\Leftrightarrow-10x-15-12+9x=13\)
=>-x-27=13
=>-x=40
hay x=-40
c: \(\Leftrightarrow-10x+70+20-5x=-15\)
\(\Leftrightarrow-15x=-105\)
hay x=7
d: \(\Leftrightarrow8x-12-7x+14=10\)
=>x+2=10
hay x=8
e: \(\Leftrightarrow-12x-18+14x+2=2\)
=>2x-16=2
hay x=9
24 tháng 9 2019
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x
30 tháng 1 2022
a: \(\Leftrightarrow x\cdot\dfrac{1}{2}=\dfrac{1}{4}\cdot\dfrac{-7}{2}+\dfrac{5}{3}=\dfrac{19}{24}\)
hay x=19/12
b: \(\Leftrightarrow\left(\dfrac{45}{11}-3x\right)\cdot\dfrac{11}{5}=\dfrac{21}{5}\)
\(\Leftrightarrow-3x+\dfrac{45}{11}=\dfrac{21}{11}\)
=>-3x=-24/11
=>x=8/11
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
d: \(\Leftrightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{4}{5}\\2x+\dfrac{3}{5}=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
câu hỏi là j