Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+y^2-z^2=x^2+\left(y-z\right)\left(y+z\right)=x^2-x\left(y-z\right)=x\left(x-y+z\right)=x\left(-y-y\right)=-2xy\)
Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)
Cộng VTV:
\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)
Để M xác định thì \(x,y,z\ne0\)
\(xy+xz+yz=0\Rightarrow\left\{{}\begin{matrix}\dfrac{xy}{z}+x+y=0\\\dfrac{xz}{y}+x+z=0\\\dfrac{yz}{x}+y+z=0\end{matrix}\right.\)
Cộng vế với vế ta được:
\(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}+2\left(x+y+z\right)=0\)
\(\Leftrightarrow M+2.\left(-1\right)=0\Rightarrow M=2\)
Ta có :
\(xy+yz+xz=0\\ \Rightarrow\left[{}\begin{matrix}xy=-xz-yz=-z\left(x+y\right)\\yz=-xy-xz=-x\left(y+z\right)\\xz=-xy-yz=-y\left(x+z\right)\end{matrix}\right.\)
\(M=\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=\dfrac{-z\left(x+y\right)}{z}+\dfrac{-y\left(x+z\right)}{y}+\dfrac{-x\left(y+z\right)}{x}\\ =-\left(x+y\right)-\left(x+z\right)-\left(y+z\right)=-x-y-x-z-y-z\\ =-2\left(x+y+z\right)=\left(-2\right)\cdot\left(-1\right)=2\)
\(\Rightarrow M=2\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
\(P=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{xy}{z}+\dfrac{zx}{y}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)\right]\)
\(\ge\dfrac{1}{2}\left(2y+2x+2z\right)=x+y+z=2014\)
Dấu = xảy ra khi \(x=y=z=\dfrac{2014}{3}\)
\(A=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{xyz+yz+y}\)
\(=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)
\(=\dfrac{xyz}{y+xyz+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)
\(=\dfrac{2019}{y+2019+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)
\(=\dfrac{yz+y+2019}{yz+y+2019}=1\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{xy}{z}+\dfrac{yz}{x}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}=2\sqrt{y^2}=2y\left(1\right)\)
\(\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}}=2\sqrt{z^2}=2z\left(2\right)\)
\(\dfrac{xy}{z}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{xz}{y}}=2\sqrt{x^2}=2x\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta được :
\(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\) ≥ \(2\left(x+y+z\right)\)
⇔ \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(x+y+z=2019\)
⇒ \(P_{Min}=2019\) ⇔ \(x=y=z=673\)
Dụng cosi để tìm GTNN hoặc GTLN nha