\(|x|,|y|,|z|\)luôn \(\ge0\forall x,y,z\)
\(\Rightarrow|x|+|y|+|z|\ge0\)
mà \(|x|+|y|+|z|\le0\left(gt\right)\)
\(\Rightarrow|x|+|y|+|z|=0\)\(\Leftrightarrow x=y=z=0\)
Vậy \(x=y=z=0\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(|x|,|y|,|z|\)luôn \(\ge0\forall x,y,z\)
\(\Rightarrow|x|+|y|+|z|\ge0\)
mà \(|x|+|y|+|z|\le0\left(gt\right)\)
\(\Rightarrow|x|+|y|+|z|=0\)\(\Leftrightarrow x=y=z=0\)
Vậy \(x=y=z=0\)
Tìm \(x\in Z\) biết:
\(\left|5x-2\right|\le0\)
Chứng minh rằng:
\(\left(y-z\right)^3.\left(1-x^3\right)+\left(z-x\right)^3.\left(1-y^3\right)+\left(x-y\right)^3.\left(1-z^3\right)=3\left(1-xyz\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
Hãy so sánh 2 biểu thức sau , biết rằng \(x;y\in Z\):
\(\left[\left(x+y\right)\left(x+y\right)+\left(x-y\right)\left(x-y\right)\right]\left[\left(x+y\right)\left(x+y\right)-\left(x-y\right)\left(x-y\right)\right]\)
và
\(\left[\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x-y\right)\right]\left[\left(x+y\right)\left(x-y\right)-\left(x+y\right)\left(x-y\right)\right]\)
1. \(\left(1-x\right)^2+\left(3-y\right)^2+\left(y^2-x-z\right)^2=0\)
2. \(\left(x-y+z^2\right)+\left(y-2\right)^2+\left(z+3\right)^2=0\)
Làm hộ mình 2 câu này
Tìm x,y,z biết \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
tìm x,y,z biết \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
Timd x,y,z biết \(\left(x-y-z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
giúp mình nhanh nha
tìm x,y,z thuộc Q biết
\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
Chứng minh đẳng thức
a) \(\left(x-y\right)-\left(x-z\right)=\left(z+x\right)-\left(y+x\right)\)
b) \(\left(x-y+z\right)-\left(y+z-x\right)-\left(x-y\right)=\left(z-y\right)-\left(z-x\right)\)
c) \(a\left(b+c\right)-b\left(a-c\right)=\left(a+b\right)c\)
d) \(a\left(b-c\right)-a\left(b+d\right)=-a\left(c+d\right)\)
e) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)=\left(a-c\right)\left(d-b\right)\)
f) \(\left(a-c\right)\left(b+d\right)-\left(a-d\right)\left(b+c\right)=\left(a+b\right)\left(d-c\right)\)