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1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
a) \(P=\left(\frac{2\sqrt{2}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3}{\sqrt{x}+3}\)
\(\frac{x^2+4x+4}{x}\left(+\right)=x+4+\frac{4}{x}Ap-dung-bdt-co-si--TA-CO\\ x+\frac{4}{x}\ge2\sqrt{x\cdot\frac{4}{x}}=4=>\left(+\right)\ge4+4=8\\ Dau-"="-xay-ra-tai-x=2\)
b/ Ko biết yêu cầu
4/ \(E=\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{x^3}+\frac{1}{x^3}\ge5\sqrt[5]{\frac{x^6}{27x^6}}=\frac{5}{\sqrt[5]{27}}\)
Dấu "=" xảy ra khi \(\frac{x^2}{3}=\frac{1}{x^3}\Leftrightarrow x=\sqrt[5]{3}\)
\(F=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge3\sqrt[3]{\frac{x^2}{4x^2}}=\frac{3}{\sqrt[3]{4}}\)
Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{1}{x^2}\Rightarrow x=\sqrt[3]{2}\)
6/ \(Q=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{8\left(x+1\right)}{2\left(x+1\right)}}=4\)
Dấu "=" xảy ra khi \(\frac{x+1}{2}=\frac{8}{x+1}\Leftrightarrow x=3\)
7/
\(R=\frac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}=10\)
Dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)
8/
\(S=x^2+\frac{2000}{x}=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{1000^2x^2}{x^2}}=300\)
Dấu "=" xảy ra khi \(x^2=\frac{1000}{x}\Leftrightarrow x=10\)