2/3 - x = 3x + 1,5
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a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
b) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
\(\frac{1}{2}.\left(x-\frac{2}{3}\right)-20\%.\left(3x-1,5\right)=0,4\)
\(\Rightarrow\frac{1}{2}.\left(x-\frac{2}{3}\right)-\frac{1}{5}.\left(3x-\frac{3}{2}\right)=\frac{2}{5}\)
\(\Rightarrow\frac{1}{2}.x-\frac{1}{3}-\frac{3}{5}x-\frac{3}{10}=\frac{2}{5}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{3}{5}\right).x=\frac{2}{5}+\frac{3}{10}+\frac{1}{3}\)
\(\Rightarrow\frac{-1}{10}.x=\frac{31}{30}\)
\(\Rightarrow x=\frac{31}{30}:\frac{-1}{10}\)
\(\Rightarrow x=-\frac{31}{3}\)
P/s : Tính theo cách này đúng nhưng sao mk thử lại sai, có j sai trong câu trl mong cacau giúp đỡ :v
\(\frac{1}{2}\left[x-\frac{2}{3}\right]-20\%\left[3x-\frac{1}{5}\right]=0,4\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}-\frac{20}{100}\left[3x-\frac{1}{5}\right]=0,4\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}-\frac{1}{5}\left[3x-\frac{1}{5}\right]=0,4\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}-\frac{3x}{5}-\frac{1}{25}=0,4\)
Làm nốt :v
`@` `\text {Ans}`
`\downarrow`
`a)`
\(2x(4x²+x-3)\)
`= 2x*4x^2 + 2x*x + 2x*(-3)`
`= 8x^3 + 2x^2 - 6x`
`b)`
\([(-9x)+4x]: (-3x)-3x²\)
`= (-9x) \div (-3x) + 4x \div (-3x) - 3x^2`
`= 3 - 4/3 - 3x^2`
`= 5/3 - 3x^2`
\(2\left|x-1\right|-7=-3\)
\(2\left|x-1\right|=4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)
\(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
a, | 5 - 3x | + \(\frac{2}{3}=\frac{1}{6}\)
=> | 5 - 3x | = \(\frac{1}{6}-\frac{2}{3}\)
=> | 5 - 3x | = \(-\frac{1}{2}\)( vô lý , vì | 5 - 3x | \(\ge\)0 )
Vậy không có giá trị của x
b, - 2,5 + | 3x + 5 | = - 1,5
=> | 3x + 5 | = - 1,5 + 2,5
=> | 3x + 5 | = 1
=> \(\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}\Rightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}}\)
Vậy x = -4 / 3 hoặc x = -2
c, \(\frac{11}{5}-\left|\frac{1}{5}-x\right|=\frac{3}{5}\)
=> \(\left|\frac{1}{5}-x\right|=\frac{11}{5}-\frac{3}{5}\)
=> \(\left|\frac{1}{5}-x\right|=\frac{8}{5}\)
=> \(\orbr{\begin{cases}\frac{1}{5}-x=\frac{8}{5}\\\frac{1}{5}-x=-\frac{8}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{5}\\x=\frac{9}{5}\end{cases}}}\)
Vậy x = - 7 / 5 hoặc x = 6 / 5
d, \(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\)
=> \(\left|x-\frac{2}{5}\right|=\frac{3}{4}-\frac{1}{2}\)
=> \(\left|x-\frac{2}{5}\right|=\frac{1}{4}\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}}\)
Vậy x = 13 / 20 hoặc x = 3 / 20
\(1,5x-2\frac{1}{3}x=1,5-\frac{2}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{7}{3}x=\frac{3}{2}-\frac{2}{3}\)
\(\Leftrightarrow-\frac{5}{6}x=\frac{5}{6}\)
\(\Leftrightarrow x=-1\)
\(1,5x-2\frac{1}{3}x=1,5-\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{3}{2}-\frac{7}{3}\right)x=\frac{5}{6}\)
\(\Leftrightarrow-\frac{5}{6}x=\frac{5}{6}\)\(\Rightarrow x=-1\)