Tìm x biết:
\(\left|2x-3\right|-x=\left|2-x\right|\)
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25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
10 tháng 3 2022
a)TH1: \(2x-3>0;3x+2>0\)
\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)
TH2: \(2x-3< 0;3x+2< 0\)
\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)
Cả 2 TH ra \(x=-5=>x=-5\)
b)TH1 \(\dfrac{1}{2}x>0\)
\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)
TH2 \(\dfrac{1}{2}x< 0\)
\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)
\(=>x=2;\dfrac{6}{5}\)
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
23 tháng 1 2024
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
23 tháng 6 2017
\(x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(x^3-25x-x^3-8=3\)
\(-25x-8=3\)
\(-25x=3+8=11\)
\(x=\frac{-11}{25}\)