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Áp dụng BĐT BSC và BĐT Cosi:
\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge17\left(x+y+z\right)+\frac{2.\left(1+1+1\right)^2}{x+y+z}\)
\(=17\left(x+y+z\right)=\frac{18}{x+y+z}\)
\(=17\left(x+y+z\right)=\frac{17}{x+y+z}+\frac{1}{x+y+z}\)
\(\ge2\sqrt{17\left(x+y+z\right).\frac{17}{x+y+z}}+\frac{1}{1}\)
\(=35\)
\(\Rightarrow17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Áp dụng bất đẳng thức AM-GM kết hợp giả thiết x + y + z ≤ 1 ta có :
\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=17x+17y+17z+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)
\(=\left(18x+\frac{2}{x}\right)+\left(18y+\frac{2}{y}\right)+\left(18z+\frac{2}{z}\right)-\left(x+y+z\right)\)
\(\ge2\sqrt{18x\cdot\frac{2}{x}}+2\sqrt{18y\cdot\frac{2}{y}}+2\sqrt{18z\cdot\frac{2}{z}}-1=12\cdot3-1=35\)( đpcm )
Dấu "=" xảy ra <=> x=y=z=1/3
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)
c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)
\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
* Nếu x + y + z = 0
\(A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\dfrac{x+y}{x}\cdot\dfrac{y+z}{y}\cdot\dfrac{z+x}{z}=\dfrac{\left(-z\right)}{x}\cdot\dfrac{\left(-x\right)}{y}\cdot\dfrac{\left(-y\right)}{z}=\dfrac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-\dfrac{xyz}{xyz}=-1\)
* Nếu x + y + z khác 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-y-z}{x}=\dfrac{y-x-z}{y}=\dfrac{-x-y+z}{z}=\dfrac{x-y-z+y-x-z-x-y+z}{x+y+z}=\dfrac{-x-y-z}{x+y+z}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-z=-x\\y-x-z=-y\\-x-y+z=-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\Rightarrow x=y=z\)
\(\Rightarrow A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
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\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
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\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
Dễ thì giải đi bn
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