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Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)
Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)
\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)
Với mọi a;b;c không âm ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)
Áp dụng:
a.
\(VT\le\sqrt{3\left(x+7+y+7+z+7\right)}=\sqrt{3\left(6+21\right)}=9\)
Dấu "=" xảy ra khi \(x=y=z=2\)
b.
\(VT\le\sqrt{3\left(3x+2y+3y+2z+3z+2x\right)}=\sqrt{15\left(x+y+z\right)}=\sqrt{15.6}=3\sqrt{10}\)
Dấu "=" xảy ra khi \(x=y=z=2\)
c.
\(VT\le\sqrt{3\left(2x+5+2y+5+2z+5\right)}=\sqrt{3\left(2.6+15\right)}=9\)
Dấu "=" xảy ra khi \(x=y=z=2\)
\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
\(6=\dfrac{1}{x}+\dfrac{4}{2y}+\dfrac{9}{3z}\ge\dfrac{\left(1+2+3\right)^2}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
Đề câu g có vấn đề aa :>>>
Câu còn lại tương tự như trên mà
