Tìm x,y (x-0,5)^2+(y+0,25)^2=0
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Ta có:
\(\left(x-0,5\right)^2+\left(y+0,25\right)^2=0\)
\(\Leftrightarrow x-0,5=y+0,25=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=-0,25\end{matrix}\right.\)
0,25 x y = 0,125 + 0,5
0,25 x y = 0,625
y = 0,625 : 0,25
y = 0,25
vậy x = ...
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) \(\frac{1}{5}.\frac{x}{3}=\frac{2}{3}:0,25\)
\(=>\frac{x}{15}=\frac{4}{3}\)
\(=>x=\frac{4.15}{3}=20\)
1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a ) y : 0.5 + y x 3 = 2,3
y x 2 + y x 3 = 2,3
y x ( 2 + 3 ) = 2,3
y x 5 = 2,3
y =0,46
b ) y : 0 , 25 - y : 2,5 = 0,18
y x 4 - y x \(\frac{2}{5}\) = 0,18
y x ( 4 - \(\frac{2}{5}\)) = 0 , 18
y x 3,6 = 0 , 18
y =0,05
y : 0,5 + y : 0,25 + y x 4 = 106,7
y x 2 + y x 4 + y x 4 = 106,7
y x (2+4+4)= 106,7
y x 10= 106,7
y= 106,7:10
y= 10,67
\(\left(x-0,5\right)^2+\left(y+0,25\right)^2=0\)
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 \(\left(x-0,25\right)^2+\left(y-0,25\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+0,5\right)^2=0\\\left(y-0,25\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+0,5=0\\y-0,25=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-0,5\\y=0,25\end{cases}}\)