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a: =>-0,5x+1,5=0,4x-0,2
=>-0,9x=-1,7
=>x=17/9
3x-1/2x+3=3x+2/2x-1
=>6x^2-3x-2x+1=6x^2+4x+9x+6
=>-5x+1=13x+6
=>-8x=5
=>x=-5/8
b: \(\Leftrightarrow\left(4x-1\right)\left(-x+7\right)=\left(4x+5\right)\left(-x-2\right)\)
=>\(-4x^2+28x+x-7=-4x^2-8x-5x-10\)
=>29x-7=-13x-10
=>42x=-3
=>x=-1/14
c: =>7x=5y và 2x-y=15
=>7x-5y=0 và 2x-y=15
=>x=25; y=35
1) ADTCDTSBN, ta có:
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4
* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12
- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16
* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20
Vậy x = 12
y = 16
z = 20
Ta có : |3x - 2| - 1 = x
=> |3x - 2| = x + 1
\(\Leftrightarrow\orbr{\begin{cases}3x-2=x+1\\3x-2=-x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-x=1+2\\3x+x=-1+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=3\\4x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{4}\end{cases}}\)
x) \(\left|2x-5\right|=x+1\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=1+5\\2x+x=-1+5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\3x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
y) \(\left|3x-2\right|-1=x\)
\(\Leftrightarrow\left|3x-2\right|=x+1\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-x=1+2\\3x+x=-1+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\4x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)
z) \(\left|3x-7\right|=2x+1\)
\(\Rightarrow\left[{}\begin{matrix}3x-7=2x+1\\3x-7=-2x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2x=1+7\\3x+2x=-1+7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\5x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{6}{5}\end{matrix}\right.\)
v) \(\left|2x-1\right|+1=x\)
\(\Leftrightarrow\left|2x-1\right|=x-1\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=x-1\\2x-1=-x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=-1+1\\2x+x=1+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)