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\(y=-x^2+40x+600\)
\(=-\left(x^2-40x+400\right)+1000\)
\(-\left(x-20\right)^2+1000\le1000\)
\(y_{max}=1000\Leftrightarrow x=20\)
Bài 2:
Ta có: M = a2+ab+b2 -3a-3b-3a-3b +2001
=> 2M = ( a2 + 2ab + b2) -4.(a+b) +4 + (a2 -2a+1)+(b2 -2b+1) + 3996
2M= ( a+b-2)2 + (a-1)2 +(b-1)2 + 3996
=> MinM = 1998 tại a=b=1
Câu 3:
Ta có: P= x2 +xy+y2 -3.(x+y) + 3
=> 2P = ( x2 + 2xy +y2) -4.(x+y) + 4 + (x2 -2x+1) +(y2 -2y+1)
2P = ( x+y-2)2 +(x-1)2+(y-1)2
=> MinP = 0 tại x=y=1
\(Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)
\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+xy+yz+zx\right)}+\sqrt{6\left(y^2+xy+yz+zx\right)}+\sqrt{z^2+xy+yz+zx}}\)
\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{3\left(x+y\right).2\left(x+z\right)}+\sqrt{3\left(y+x\right).2\left(y+z\right)}+\sqrt{\left(z+x\right).\left(z+y\right)}}\)
\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(y+x\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}}\)
\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{9x+9y+6z}{2}}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)và \(z=2\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
K = a² - 2ab + 5b² - 4b + 9
= (a² - 2ab + b²) + (4b² - 4b + 1) + 8
= (a - b)² + (2b - 1)² + 8
Do (a - b)² ≥ 0 với mọi a, b ∈ R
(2b - 1)² ≥ 0 với mọi b R
⇒ (a - b)² + (2b - 1)² ≥ 0 với mọi a, b ∈ R
⇒ (a - b)² + (2b - 1)² + 8 ≥ 8 với mọi a, b ∈ R
Vậy GTNN của K là 8 khi a = b = 1/2