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Ta có
\(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(z+1\right)^2\ge0\end{cases}}\)và \(\hept{\begin{cases}x^2+1>0\\y^2+1>0\\z^2+1>0\end{cases}}\)
\(\Rightarrow A=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}\ge0\)
Kết hợp với điều kiện ban đầu thì
GTNN của A là 0 đạt được khi
\(\left(x,y,z\right)=\left(-1,-1,5;-1,5,-1;5,-1-1\right)\)
Anh có cách khác nè :
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{yz\left(x-y+z-x\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(yz-xy\right)-\left(z-x\right)\left(zx-yz\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{y\left(x-y\right)\left(z-x\right)-z\left(x-y\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-x\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2z+yz^2-z^2x+zx^2-x^2y+xy^2}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2\left(z-x\right)-zx\left(z-x\right)+y\left(z^2-x^2\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-x\right)\left(yz+xy-y^2-zx\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-y\right)\left[y\left(x-y\right)-z\left(x-y\right)\right]}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
Bạn đã ib nhờ mik thì mik làm cho trót vại UwU
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-x\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}.\)
\(=-\frac{1}{x\left(x-y\right)\left(z-x\right)}-\frac{1}{y\left(y-z\right)\left(x-y\right)}-\frac{1}{z\left(z-x\right)\left(y-z\right)}\)
\(=-\frac{y^2x-yz^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{xz^2-x^2z}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{x^2y-xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-y^2z+yz^2-xz^2+x^2z-x^2y+xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-\left(y^2z-x^2z\right)+\left(yz^2-xz^2\right)-\left(x^2y-xy^2\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-z\left(y^2-x^2\right)+z^2\left(y-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-z\left(y-x\right)\left(x+y\right)+z^2\left(y-x\right)+xy\left(y-x\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{-xyz\left(y-x\right)\left(z-x\right)\left(y-z\right)}\)
\(=-\frac{-z\left(x+y\right)+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=-\frac{-zx-zy+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-\left(zx-xy\right)-\left(zy-z^2\right)}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=\frac{-x\left(z-y\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=\frac{x\left(y-z\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)
\(=\frac{x-z}{xyz\left(z-x\right)}\)
\(=\frac{-\left(z-x\right)}{xyz\left(z-x\right)}\)
\(=\frac{-1}{xyz}\)
Câu 2/
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}=1\)
Điều kiện \(\hept{\begin{cases}x^2\ne0\\x^2+y^2\ne0\\x^2+y^2+z^2\ne0\end{cases}}\)
Xét \(x^2,y^2,z^2\ge1\)
Ta có: \(\hept{\begin{cases}x^2\ge1\\x^2+y^2\ge2\end{cases}}\)
\(\Rightarrow x^2\left(x^2+y^2\right)\ge2\)
\(\Rightarrow\frac{1}{x^2\left(x^2+y^2\right)}\le\frac{1}{2}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}\le\frac{1}{6}\left(2\right)\\\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{3}\left(3\right)\end{cases}}\)
Cộng (1), (2), (3) vế theo vế ta được
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{2}+\frac{1}{6}+\frac{1}{3}=1\)
Dấu = xảy ra khi \(x^2=y^2=z^2=1\)
\(\Rightarrow\left(x,y,z\right)=?\)
Xét \(\hept{\begin{cases}x^2\ge1\\y^2=z^2=0\end{cases}}\) thì ta có
\(\frac{1}{x^4}+\frac{1}{x^4}+\frac{1}{x^4}=1\)
\(\Leftrightarrow x^4=3\left(l\right)\)
Tương tự cho 2 trường hợp còn lại: \(\hept{\begin{cases}x^2,y^2\ge1\\z^2=0\end{cases}}\) và \(\hept{\begin{cases}x^2,z^2\ge1\\y^2=0\end{cases}}\)
Bài 2/
Ta có: \(\frac{x}{y}+\frac{y}{z}+\frac{z}{t}+\frac{t}{x}\ge4\sqrt[4]{\frac{x}{y}.\frac{y}{z}.\frac{z}{t}.\frac{t}{x}}=4>3\)
Vậy phương trình không có nghiệm nguyên dương.
\(\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right).\left(1+\frac{1}{z}\right)=2\)
Giả sử \(x\ge y\ge z>0\)
\(\Rightarrow\frac{1}{x}\le\frac{1}{y}\le\frac{1}{z}\)
\(\Rightarrow1+\frac{1}{x}\le1+\frac{1}{y}\le1+\frac{1}{z}\)
\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\le \left(1+\frac{1}{z}\right)^3\)
\(\Rightarrow2\le\left(1+\frac{1}{z}\right)^3\)
\(\Rightarrow1+\frac{1}{z}\ge\sqrt[3]{2}\)
\(\Rightarrow\frac{1}{z}\ge\sqrt[3]{2}-1\)
\(\Rightarrow z\le\frac{1}{\sqrt[3]{2}-1}< 4\)
Mà z thuộc N* \(\Rightarrow z\in\left\{1;2;3\right\}\)
TH1 : \(z=1\)
\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{1}\right)=2\)
\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=1\)
Ta có : \(1+\frac{1}{x}>1;1+\frac{1}{y}>1\)\(\Rightarrow\left(\frac{1}{x}+1\right)\left(1+\frac{1}{y}\right)>1\left(lọai\right)\)
TH2 : \(z=2\)
\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{2}\right)=2\)
\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)
Ta có : \(\left(1+\frac{1}{y}\right)^2\ge\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)
\(\Rightarrow1+\frac{1}{y}\ge\sqrt{\frac{4}{3}}\)
\(\Rightarrow\frac{1}{y}\ge\frac{2\sqrt{3}}{3}-1\)
\(\Rightarrow y\le\frac{1}{\frac{2\sqrt{3}}{3}-1}< 7\)
\(\Rightarrow y\in\left\{1;2;3;4;5;6\right\}\)
Nếu y = 1 \(\Rightarrow\left(1+1\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > x = -3 ( loại )
Nếu y = 2 \(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > x = -9 ( loại )
Nếu y = 3 \(\Rightarrow\left(1+\frac{1}{3}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > \(x\in\varnothing\)
Nếu y = 4 \(\Rightarrow\left(1+\frac{1}{4}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > x = 15 ( tm )
Nếu y = 5 \(\Rightarrow\left(1+\frac{1}{5}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > x = 9 ( tm )
Nếu y = 6 \(\Rightarrow\left(1+\frac{1}{6}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)
= > x = 7 ( tm )
TH3 : z =3 thì bạn làm tương tự nhé
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