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Chứng minh rằng:
\(\frac{\left|x\right|}{2008+\left|x\right|}+\frac{\left|y\right|}{2008+\left|y\right|}\ge\frac{\left|x-y\right|}{2008+\left|x-y\right|}\)
P/S: - Việt ~~ Giúp tttttttttttttttttttttttttttttttttt
Khai triển và thu gọn 2 đa thức \(f\left(x\right)=\left(x-2\right)^{2008}+\left(2x-3\right)^{2007}+2006x\) và \(g\left(x\right)=y^{2009}-2007y^{2008}+2005y^{2007}\)
Tìm x,y,z
\(\left|x-3\right|+\left|y-2x\right|+\left|2z-x+y\right|=0\)
\(\left|x-y\right|+\left|2y+x-\frac{1}{2}\right|+\left|x+y+z\right|\le0\)
Tìm x,y,za)frac{x}{4}-frac{1}{9}frac{1}{2}left(xthuộcZright)b)x+yxyx:yleft(vớiright)xykhác0c)x-yxyxyleft(ykhac0right)d)left(x+1right)left(x-2right) 0e)left(x-2right)left(x+frac{2}{3}right)0f)xleft(x+y+zright)-5yleft(x+y+zright)9zleft(x+y+zright)5
Đọc tiếp
Tìm x,y,z
a)\(\frac{x}{4}-\frac{1}{9}=\frac{1}{2}\left(xthuộcZ\right)\)
b)\(x+y=xy=x:y\left(với\right)xykhác0\)
c)\(x-y=xy=xy\left(ykhac0\right)\)
d)\(\left(x+1\right)\left(x-2\right)< 0\)
e)\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
f)\(x\left(x+y+z\right)=-5\)
\(y\left(x+y+z\right)=9\)
\(z\left(x+y+z\right)=5\)
Tìm số hữu tỉ x, y biết: \(\left(3x-33\right)^{2008}+\left|y-7\right|^{2009}\le0\)
a, A= \(\left(15x+2y\right)-\left[\left(2x+3\right)-\left(5x+y\right)\right]\) tại x=1 ; y= -1
b, B= -\(\left(12x+3y\right)+\left(5x-2y\right)-\left[13x+\left(2y+5\right)\right]\) tại x= \(\frac{-1}{2}\); y= \(\frac{1}{7}\)
1 tìm x,y biết\(a,\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
Xem chi tiết
\(b.\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
a) Tìm x,y,z biết : \(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)
b) Tìm x biết : \(\left|x\right|+\left|x-1\right|+\left|x-2\right|=6\)
\(\left|x-\frac{1}{2}\right|+\left|x+\right|y-\frac{1}{2}\left|\right|=0\)