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Ta co:\(\Sigma\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}=\Sigma\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)Ta lai co:
\(\Sigma x+\Sigma\frac{1}{x}=\Sigma\left(x+\frac{1}{4x}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+\frac{3}{4}.\frac{9}{x+y+z}\ge3+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{2}\)
Vay \(P_{min}=\frac{15}{2}\)khi \(x=y=z=\frac{1}{2}\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
\(x=\sqrt{\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}}\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)}\)
\(=\sqrt{2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=\sqrt{2}\)
2/ Để đồ thị hàm số cắt 2 trục tọa độ tại 2 điểm pb \(\Leftrightarrow\left\{{}\begin{matrix}m-1\ne0\\m\ne0\end{matrix}\right.\)
Gọi A là giao điểm của (d) với trục Ox \(\Rightarrow A\left(\frac{2m}{1-m};0\right)\)
\(\Rightarrow OA=\left|\frac{2m}{1-m}\right|=\left|\frac{2m}{m-1}\right|\)
Gọi B là giao điểm của (d) với Oy \(\Rightarrow B\left(0;2m\right)\Rightarrow OB=\left|2m\right|\)
\(S_{OAB}=\frac{1}{2}OA.OB=1\Leftrightarrow OA.OB=2\)
\(\Leftrightarrow\left|\frac{2m}{m-1}\right|.\left|2m\right|=2\Leftrightarrow2m^2=\left|m-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2m^2=m-1\\2m^2=1-m\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2m^2-m+1=0\left(vn\right)\\2m^2+m-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=-1\\m=\frac{1}{2}\end{matrix}\right.\)
3/
a/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left(x+3\right)\sqrt{x+3}+2\sqrt{x+3}=\left(x+1\right)\left[\left(x+1\right)^2+2\right]\)
\(\Leftrightarrow\left(x+3\right)\sqrt{x+3}+2\sqrt{x+3}=\left(x+1\right)^3+2\left(x+1\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+3}=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow a^3+2a=b^3+2b\)
\(\Leftrightarrow a^3-b^3+2\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+2\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+2\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow\sqrt{x+3}=x+1\) (\(x\ge-1\))
\(\Leftrightarrow x+3=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+x-2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)