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\(\left(0,5-2x\right)\dfrac{3}{2}=-3\)
\(\Rightarrow\left(\dfrac{1}{2}-2x\right)\dfrac{3}{2}=-3\)
\(\Rightarrow\left(\dfrac{1}{2}-2x\right)=-3.\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{2}-2x=-2\)
\(\Rightarrow2x=\dfrac{1}{2}+2\)
\(\Rightarrow2x=\dfrac{5}{2}\)
\(\Rightarrow x=\dfrac{5}{2}.\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{5}{4}\)
0,5x-2/3(x-1)=1/6
1/2x-(2/3x-2/3)=1/6
1/2x-2/3x+2/3=1/6
x.(1/2-2/3)=1/6-2/3
x.(-1/6)=-1/2
x.(-1/6)=-1/2
x=-1/2:(-1/6)
x=-1/2.(-6)
x=3
Vậy x=3
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(\Leftrightarrow-\frac{1}{4}x=\frac{4}{9}\)
\(\Leftrightarrow x=-\frac{16}{9}\)
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(< =>\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(< =>-\frac{1}{4}x=\frac{4}{9}\)
\(=>x=\frac{-16}{9}\)
a) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|2x\right|-2,5=7,5\)
\(\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\begin{cases}x\ge0\\2x=10\end{cases}\) hoặc \(\begin{cases}x< 0\\2x=-10\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=5\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-5\left(tm\right)\end{cases}\)
Vậy x={5;-5}
b)\(\left|3x\right|\cdot\left|-3,5\right|=\left|-2,8\right|\)
\(\Leftrightarrow\left|3x\right|\cdot3,5=2,8\)
\(\Leftrightarrow\left|3x\right|=\frac{4}{5}\)
\(\Leftrightarrow\begin{cases}x\ge0\\3x=\frac{4}{5}\end{cases}\) hoặc \(\begin{cases}x< 0\\3x=-\frac{4}{5}\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=\frac{4}{15}\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-\frac{4}{15}\end{cases}\)
Vậy x={4/15;-4/15}
c) \(\left(3x-5\right)\left(\frac{3}{2}x+2\right)\left(0,5x-10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-5=0\\\frac{3}{2}x+2=0\\0,5x-10=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{4}{3}\\x=20\end{array}\right.\)
a)|2x|-|-2,5|=|-7,5|
|2x|-2,5=7,5
|2x|=10
\(\Rightarrow\left[\begin{array}{nghiempt}2x=10\\2x=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
Vậy x=5;-5
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)ĐKXĐ: \(x\ne-\frac{1}{2};x\ne-3\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+x+3x+3=x^2+0,5x+4x+2\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow x^2-x^2+4x-4,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\)
\(\Leftrightarrow x=2\)
vậy x=2