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a) ta có:
\(|2x-6|+5x=9\Leftrightarrow|2x-6|=9-5x\)
\(2x-6=9-5x\Leftrightarrow7x=15\Leftrightarrow x=\frac{15}{7}\)
\(2x-6=5x-9\Leftrightarrow3x=3\Leftrightarrow x=1\)
b) Ta có:
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)nên \(x+329=0\Leftrightarrow x=-329\)
Vậy ............................................. chúc bn hok tốt ^-^
\(\left(3x+2\right)-\left(x-1\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x+2-x+1=4x+4\)
\(\Leftrightarrow3x+2-x+1-4x-4=0\)
\(\Leftrightarrow\left(3x-x-4x\right)+\left(1+2-4\right)=0\)
\(\Leftrightarrow-2x-1=0\)
\(\Leftrightarrow-2x=0+1\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(\left|x-1\right|+1=2x-3\)
\(\Leftrightarrow\left|x-1\right|=2x-3-1\)
\(\Leftrightarrow\left|x-1\right|=2x-4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x-4\\x-1=4-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{3};3\right\}\)
a) 3x - / 2x + 1/=2
Ta co: /2x+1/ lon hon hoac bang 0
ma 3x- / 2x+1/ = 2
=> 3x la so tu nhien
=>3x-/2x+1/ = 3x - 2x+1 = 2
=>3x - 2x = 1
=>x(3-2) = 1
=>x . 1 = 1
=> x=1
KL........\
Tich cho minh nhe ! Cau b dang suy nghi .
a) Ta co: /2x+1/ lon hon hoac bang 0
ma 3x - /2x+1/ = 2
=> 3x la so tu nhien
=> 3x - /2x+1/ = 3x -2x +1 = 2\
=> 3x -2x =1
=>x=1
tick cho minh nha!!!!! Thank you nhieuuuuuuuuu !!!!