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a: ta có: \(2\left(4-3x\right)+2x=5\left(2x-3\right)\)
\(\Leftrightarrow8-6x+2x-10x+15=0\)
\(\Leftrightarrow-14x=-23\)
hay \(x=\dfrac{23}{14}\)
b: Ta có: \(\dfrac{1}{2}-\left(2x-\dfrac{1}{3}\right)^2=\dfrac{7}{18}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{3}\\2x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{2}{3}\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)
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\(\left(2x+1\right)^3=18\)
\(\Rightarrow2x+1=\sqrt[3]{18}\)
\(\Rightarrow x=\frac{\sqrt[3]{18}-1}{2}\)
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\((2x+3)-13\times40=-18\\\Rightarrow (2x+3)-520=-18\\\Rightarrow2x+3=-18+520\\\Rightarrow 2x+3=502\\\Rightarrow2x=502-3\\\Rightarrow2x=499\\\Rightarrow x=\dfrac{499}{2}\\Vậy:x=\dfrac{499}{2}\)
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`4|8x-3|+5|3-2x|=18`
`<=>4|8x-3|+5|2x-3|=18`
Nếu `x>=3/2=>|8x-3|=8x-3,|2x-3|=2x-3`
`pt<=>4(8x-3)+5(2x-3)=18`
`<=>32x-12+10x-15=18`
`<=>42x=27+18=45`
`<=>x=45/42(l)`
Nếu `x<=3/8=>|8x-3|=3-8x,|2x-3|=3-2x`
`pt<=>4(3-8x)+5(3-2x)=18`
`<=>12-32x+15-10x=18`
`<=>27-42x=18`
`<=>9=42x`
`<=>x=3/14(tm)`
Nếu `3/9<=x<=3/2=>|8x-3|=8x-3,|2x-3|=3-2x`
`pt<=>4(8x-3)+5(3-2x)=18`
`<=>32x-12+15-10x=18`
`<=>22x+3=18`
`<=>22x=15`
`<=>x=15/22(tm)`
Vậy x=15/22 hoặc x=3/14
\(\frac{3}{18}=\frac{1}{2x}\)
=> \(3\cdot2x=18\cdot1\)
=> \(6x=18\)
=> \(x=18:6=3\)
Vậy x = 3
Ta có : \(\frac{3}{18}\)= \(\frac{1}{2x}\)
\(\Rightarrow\)3.2x = 1.18
\(\Rightarrow\)6x = 18
\(\Rightarrow\)x = 18 : 6
\(\Rightarrow\)x = 3