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a) \(2,5:4x=0,5:0,2\)
\(2,5:4x=\frac{5}{2}\)
\(4x=2,5:\frac{5}{2}\)
\(4x=1\)
\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
b) \(\frac{1}{5}.x:3=\frac{2}{3}:0,25\)
\(\frac{1}{5}.x:3=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}.3\)
\(\frac{1}{5}.x=8\)
\(x=8:\frac{1}{5}\)
\(x=40\)
Vậy \(x=40\)
a) \(\frac{2,5}{4x}=\frac{0,5}{0,2}\)
\(=>4x=\frac{0,2.2,5}{0,5}=1\)
\(=>x=\frac{1}{4}\)
b) \(\frac{1}{5}.\frac{x}{3}=\frac{2}{3}:0,25\)
\(=>\frac{x}{15}=\frac{4}{3}\)
\(=>x=\frac{4.15}{3}=20\)
\(x-0,5=-0,25\)
\(x=0,5+\left(-0,25\right)\)
\(x=0,5=0,25=0,25\)
Ta có:
\(\left(x-0,5\right)^2+\left(y+0,25\right)^2=0\)
\(\Leftrightarrow x-0,5=y+0,25=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=-0,25\end{matrix}\right.\)
\(\left(x-0,5\right)^2+\left(y+0,25\right)^2=0\)
Do \(\hept{\begin{cases}\left(x-0,5\right)^2\ge0\\\left(y+0,25\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-0,5=0\\y+0,25=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=-0,25\end{cases}}}\)
Vậy \(\hept{\begin{cases}x=0,5\\y=-0,25\end{cases}}\)
vì \(\hept{\begin{cases}\left(x-0,5\right)^2\ge0\\\left(y+0,25\right)\ge0\end{cases}}\)
mà \(\left(x-0,25\right)^2+\left(y-0,25\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+0,5\right)^2=0\\\left(y-0,25\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+0,5=0\\y-0,25=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-0,5\\y=0,25\end{cases}}\)
a) \(2,5:0,4x=0,5:0,2\)
\(\Rightarrow\frac{5}{2}:4x=\frac{1}{2}:\frac{1}{5}=\frac{5}{2}\)
\(\Rightarrow4x=\frac{5}{2}:\frac{5}{2}=1\)
\(\Rightarrow x=\frac{1}{4}\)
b) \(\frac{1}{5}x:3=\frac{2}{3}:0,25\)
\(\Rightarrow\frac{1}{5}x:3=\frac{8}{3}\)
\(\Rightarrow\frac{1}{5}x=\frac{8}{3}.3=8\Rightarrow x=40\)
a)2,5:4x=0,5:0,2
2,5:4x=2.5
4x=2,5:2,5
4x=1
x=1:4
x=0,25
(-0,25 )5 : \(x\) = (-0,25)3
\(x\) = (-0,25)5 : (-0,25)3
\(x\) = (-0,25)2
\(x\) = 0,0625
\(x=\dfrac{0,25^5}{0,25^3}=0,25^2\)