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`@` `\text {Ans}`
`\downarrow`
`a)`
`-3x - 3/4 = 6/5`
`=> -3x = 6/5 + 3/4`
`=> -3x = 39/20`
`=> x = 39/20 \div (-3)`
`=> x = -13/20`
Vậy, `x=-13/20`
`b)`
`1/7 - 3/5x = 3/5`
`=> 3/5x = 1/7 - 3/5`
`=>3/5x = -16/35`
`=> x = -16/35 \div 3/5`
`=> x = -16/21`
`c)`
`3/7 - 1/2x = 5/3`
`=> 1/2x = 3/7 - 5/3`
`=> 1/2x = -26/21`
`=> x = -26/21 \div 1/2`
`=> x = -52/21`
Vậy, `x = -52/21`
`d)`
`-2/3x + 2 = 3/4`
`=> -2/3x = 3/4 - 2`
`=> -2/3x = -5/4`
`=> x = -5/4 \div (-2/3)`
`=> x = 15/8`
Vậy, `x=15/8.`
cho : 2bx - 3cy /a= 3cx-az/2b = ay-abx/3c
chứng minh rằng : x/a=y/2b=z/3c
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).
a , \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b , \(\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)
d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)
<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
A
B