Vì (x-2)2012 ≥ 0
/y2 -9/2014 ≥ 0\
=> (x-2)2012 +/y2 -9/2014 = 0
=> (x-2)2012 = 0
/y2 - 9/ 2014 = 0
=> x-2 = 0
y2 -9 = 0
=> x = 0
y2 = 9
=> x = 0
y = 3 ; -3
\(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
Tìm x, y
giúp mình với
a)\(\left|5x+4\right|+7=26\)
b)\(3\left|9-2x\right|-17=16\)
c)\(8x-\left|4x+1\right|=x+2\)
d)\(\left|17x-5\right|-\left|17x+5\right|=0\)
e)\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|=2\)
Tìm X
\(2\left|2x-6\right|=\dfrac{5}{6}-\left|x-3\right|\)
2:\(\left|x+2013\right|+\left|x+2014\right|+\left|x+2045\right|=2\)
3:\(\left|2x-1\right|=\left|x+1\right|\)
4:\(\sqrt{\left(x+\sqrt{5}\right)}+\sqrt{\left(y-\sqrt{3}\right)^2}+\left|x-y-z\right|=0\)
1) tìm hai số x,y,z,t( nếu có) bít rằng:
+) x:y:z:t = 2:3:4:5 và x+y+z+t = - 42
+) \(\frac{x}{4}=\frac{y}{7}\) và x.y=112
+) \(\frac{x}{3}=\frac{y}{4}\) x.y=48
+) \(\frac{x}{2}=\frac{y}{-3}\) và xy = -54
+) \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
+) -0,16 : x = -x : 25
biết : \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}=0\)
vậy 4x+y-3z
a,\(\left|3x-4\right|+\left|3y+5\right|=0\) b,\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\) c,\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
d,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\) e,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)
Giups mình giải bài tìm x,y,z này nhé!!! Cảm ơn nhiều ạ!!!
Tìm x,y,z biết: \(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)
Tìm x, y, zϵ R biết: \(\left(4x^2-4x+1\right)^{2022}+\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
Tìm x,y biết:
a)\(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)
b)\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{-1}{4}-\left|y\right|\)
c)\(\left|x-y\right|+\left|\dfrac{9}{25}+y\right|=0\)