\(\left(x-3\right)^2+\left[y^2-25\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left[y^2-25\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3^2=0\\y^2-5^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=3^2\\y^2=5^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)
Vậy x=3;y=5
\(\left(x-\dfrac{1}{2}\right)^{50}+\left(y+\dfrac{1}{3}\right)^{40}=0\)
1.TÌm x,y,z biết
a.2009-\(\left|x-2009\right|\)=x
b.\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+u-z\right|\)=0
2.Tìm các số a,b,c biết
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)
và a+b+c = -50
Tìm x, y, z biết
a) \(\dfrac{x}{y+z+1}\) =\(\dfrac{y}{x+y+2}=\dfrac{z}{x+y-3}\)
b)\(6\left(x-\dfrac{1}{y}\right)=3\left(y-\dfrac{1}{2}\right)=2\left(z-\dfrac{1}{x}\right)=xyz-\dfrac{1}{xyz}\)
Giúp mik nha!
Tìm x, y, z biết \(\dfrac{-2.\left(x-3\right)}{5}=\dfrac{y+4}{-4}=\dfrac{3.\left(z-5\right)}{2}\) và x - y + z = -1
tìm x,y,z biết:
a) 4x=3y=5z và \(x^2+\left(2y-3z\right)^2=836\)
b) 2x=5y và \(\left(x-y\right)^2+\left(x-y\right)^2=50\)
Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) \(\left(a\ne0,b\ne0,c\ne0\right)\)
Chứng minh rằng: \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Tim x,y,x biet;
a/ \(x:y:z=3:5:\left(-2\right)\)va \(5.x-y+3.z\) \(=-16\)
b/ \(2.x=3.y;5.y=7.z\) va \(3.x-7.y+5.z=30\)
c/ \(x:y:z=4:5:6\) va \(x^2-2.y^2+z^2=18\)
\(\dfrac{6\left|y+5\right|+14}{2\left|y+5\right|+14}\)
Tìm x
Cho các mẫu \(\ne0\) và thỏa mãn \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\). Tính \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
