tim x, y thuoc Z : x2+102=y2
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Những câu hỏi liên quan
tim x thuoc z biet y^2-x^2=102
tim x thuoc z sao cho \(x^2+102=y^2\)
tim x, y thuoc N de :
( x + 5 ) . ( y + 2 ) = 102
c) C x(y2 +z2)+y(z2 +x2)+z(x2 +y2)+2xyz.
d) D x3(y−z)+y3(z−x)+z3(x−y).
e) E (x+y)(x2 −y2)+(y+z)(y2 −z2)+(z+x)(z2 −x2).
b) x2 +2x−24 0.
d) 3x(x+4)−x2 −4x 0.
f) (x−1)(x−3)(x+5)(x+7)−297 0.
(2x−1)2 −(x+3)2 0.
c) x3 −x2 +x+3 0.
e) (x2 +x+1)(x2 +x)−2 0.
a) A x2(y−2z)+y2(z−x)+2z2(x−y)+xyz.
b) B x(y3 +z3)+y(z3 +x3)+z(x3 +y3)+xyz(x+y+z). c) C x(y2 −z2)−y(z2 −x2)+z(x2 −y2).
Đọc tiếp
c) C = x(y2 +z2)+y(z2 +x2)+z(x2 +y2)+2xyz.
d) D = x3(y−z)+y3(z−x)+z3(x−y).
e) E = (x+y)(x2 −y2)+(y+z)(y2 −z2)+(z+x)(z2 −x2).
b) x2 +2x−24 = 0.
d) 3x(x+4)−x2 −4x = 0.
f) (x−1)(x−3)(x+5)(x+7)−297 = 0.
(2x−1)2 −(x+3)2 = 0.
c) x3 −x2 +x+3 = 0.
e) (x2 +x+1)(x2 +x)−2 = 0.
a) A = x2(y−2z)+y2(z−x)+2z2(x−y)+xyz.
b) B = x(y3 +z3)+y(z3 +x3)+z(x3 +y3)+xyz(x+y+z). c) C = x(y2 −z2)−y(z2 −x2)+z(x2 −y2).
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Cho x2+y2=(x+y-z)2
CMR x2+(x-z)2/y2+(y-z)2 =x-z/y-z
tim x, y, z thuoc Z sao cho /x-y/+/y-z/+/x-z/=2019
tim x,y,z thuoc z biet /x/+/y/+/z/=0
VÌ \(\left|x\right|\ge0;\left|y\right|\ge0;\left|z\right|\ge0\)NÊN ĐỂ\(\left|x\right|+\left|y\right|+\left|z\right|=0\)\(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\\\left|z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}}\)
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tim x, y thuoc z biet :
/ x-y / + / y-z/ + /z - x / = 2015
Cho các số x, y, z dương. Chứng minh rằng x2/y2 + y2/z2 + z2/x2 ≥ x/y + y/z + z/x