Tìm x,y,z biết:x^2yz=-2,xy^2z=2,xyz^2=-2
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x=1;y=-1;z=2 nhé bn đấy là tìm mò còn lời giải để mình nghĩ cái ( hơi lâu đấy =((( )
Ta có:
\(A=x^2yz=x.x.y.z=x.xyz\left(1\right)\)
\(B=xy^2z=x.y.y.z=y.xyz\left(2\right)\)
\(C=xyz^2=x.y.z.z=z.xyz\left(3\right)\)
Lấy (1)+(2)+(3),vế theo vế ta được:
\(A+B+C=x.xyz+y.xyz+z.xyz=\left(x+y+z\right).xyz=xyz\) (vì x+y+z=1)
Vậy A+B+C=xyz (đpcm)
Đề phải cho x;y;z dương chứ nhỉ?
Áp dụng bất đẳng thức AM-GM:
\(x^2y^2+y^2z^2\ge2\sqrt{x^2y^4z^2}=2xy^2z\)
\(y^2z^2+x^2z^2\ge2\sqrt{x^2y^2z^4}=2xyz^2\)
\(x^2y^2+x^2z^2\ge2\sqrt{x^4y^2z^2}=2x^2yz\)
Cộng theo vế:
\(2\left(x^2y^2+y^2z^2+x^2z^2\right)\ge2\left(xy^2z+x^2yz+xyz^2\right)\)
\(\Rightarrow x^2y^2+y^2z^2+z^2x^2\ge xy^2z+x^2yz+xyz^2\)
Dấu "=" khi \(x=y=z\)
hộ caiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)
\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)
\(A+B+C=x^2yz+xy^2z+xyz^2\)
\(=xyz\left(x+y+z\right)=xyz.1=xyz\)