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Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
\(\Rightarrow\)\(x+y+z=xyz\)
Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)
Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
\(Q=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)
Dấu "=" xảy ra khi A = B :
Ta được :
\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\Rightarrow\frac{x+y+z}{xyz}=0\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(N=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=3\)
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Vì \(x,y,z\) khác nhau nên \(x+y+z=0\)
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(P=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\frac{\left(-x\right)\cdot\left(-y\right)\cdot\left(-z\right)}{xyz}=-1\)
Vậy...
Ta có :x + y + z = -1 \(\Rightarrow\)x + y =-( 1 + z )
xy + yz + xz = 0 \(\Rightarrow\)xy = - z ( x + y ) = z ( z + 1 )
Tương tự : xz = y ( y + 1 ) ; yz = x . ( x + 1 )
\(M=\frac{z\left(z+1\right)}{z}+\frac{y\left(y+1\right)}{y}+\frac{x\left(x+1\right)}{x}=x+y+z+3=2\)
1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa
Bạn tham khảo tại đây:
Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)
\(\Rightarrow yz+zx+xy=0\)
Ta có : \(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-zx-xy\)
\(=x\left(x-z\right)-y\left(x-z\right)\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự : \(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz\)
\(=y\left(y-x\right)+z\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\)
\(z^2+2xy=\left(x-z\right)\left(y-z\right)\)
\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\) \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
-cách này khá dài dòng _._ (ko nghĩ đc cách ngắn hơn >: )
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow\hept{\begin{cases}-x.\left(y+z\right)=yz\\-y.\left(x+z\right)=xz\\-z.\left(x+y\right)=xy\end{cases}}\)
thay vào biểu thức P, ta có:
\(P=\left[\frac{-z.\left(y+x\right)}{z^2}+\frac{-x.\left(y+z\right)}{x^2}+\frac{-y.\left(x+z\right)}{y^2}-2\right]^{2013}\)
\(P=\left[\frac{-\left(y+x\right)}{z}+\frac{-\left(y+z\right)}{x}+\frac{-\left(x+z\right)}{y}-2\right]^{2013}\)
\(P=\left(\frac{-x^2y-xy^2-zy^2-yz^2-zx^2-xz^2}{xyz}-\frac{2xyz}{xyz}\right)^{2013}\)
\(P=\left[\left(\frac{-x^2y-zx^2}{xyz}\right)+\left(\frac{-xy^2-zy^2}{xyz}\right)+\left(\frac{-z^2y-xz^2}{xyz}\right)\right]\)
\(\text{Ta có: }-x^2y-zx^2=-x^2.\left(y+z\right),\text{mà }-x.\left(y+z\right)=yz\Rightarrow-x^2.\left(y+z\right)=xyz\)
tương tự: \(-xy^2-zy^2=xyz\text{ và }-z^2y-z^2x=xyz\)
\(\Rightarrow P=\left(\frac{3xyz-2xyz}{xyz}\right)^{2013}=1^{2013}=1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\Rightarrow x^3y^3+y^3z^3+z^3x^3=3x^2y^2z^2\) (cách cm Câu hỏi của Arthur Conan Doyle - Toán lớp 8 - Học toán với OnlineMath)
Vậy\(P=\left(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}-2\right)^{2013}=\left(\frac{x^3y^3+y^3z^3+z^3x^3}{x^2y^2z^2}-2\right)^{2013}=\left(3-2\right)^{2013}=1\)