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`@` `\text {Ans}`
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`3,8 * 2x = 1/4*8/3`
`=> 3,8*2x = 2/3`
`=> 2x = 2/3 \div 3,8`
`=> 2x = 10/57`
`=> x = 10/57 \div 2`
`=> x = 5/57`
Vậy, `x = 5/57.`
b) \(3,8:\left(2x\right)=\frac{1}{4}:\frac{8}{3}\)
= \(3,8:\left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=3,8:\frac{3}{32}=\frac{608}{15}\)
\(\Rightarrow x\frac{608}{15}:2=\frac{304}{15}\)
\(\Rightarrow\frac{304}{15}\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
\(\frac{x}{4}=\frac{3}{2}\)
\(\Rightarrow\frac{x}{4}=\frac{6}{4}\)
\(\Rightarrow x=6\)
vậy_
b)\(\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x^2=2\cdot8\)
\(x^2=16\Rightarrow x=4\)
c) \(\frac{x+3}{4}=\frac{5}{3}\)
\(3\left(x+3\right)=4\cdot5\)
\(3x+9=20\)
\(3x=11\)
\(x=\frac{11}{3}\)
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)