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a) \(A=5+\sqrt{-4x^2-4x}\)
\(A==5+\sqrt{-4x\left(x+1\right)}\)
Có: \(-4x\left(x+1\right)\le0\)
\(\Rightarrow\sqrt{-4x\left(x+1\right)}=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy: \(Max_A=5\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(B=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ: \(\hept{\begin{cases}x\ge2\\x\le4\end{cases}}\Rightarrow x\in\left\{2;3;4\right\}\)
Thay \(x=2\Rightarrow\sqrt{2-2}+\sqrt{4-2}=\sqrt{2}\)
Thay \(x=3\Rightarrow\sqrt{3-1}+\sqrt{4-3}=2\)
Thay \(x=4\Rightarrow\sqrt{4-2}+\sqrt{4-4}=\sqrt{2}\)
Vậy: \(Max_B=2\) tại \(x=3\)
Bài 2:
a)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(\ge x-1+0+3-x=2\)
Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-2=0\\x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=2\\x\le3\end{cases}}\Leftrightarrow x=2\)
Vậy MinA=2 khi x=2
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
\(E=x+\sqrt{-x^2-2x+3}\)
\(\Leftrightarrow2\sqrt{2}E=2\sqrt{2}x+2\sqrt{2}\sqrt{-x^2-2x+3}\)
\(=\left(-\left(-x^2-2x+3\right)+2\sqrt{2}\sqrt{-x^2-2x+3}-2\right)+\left(-x^2+2.x.\left(\sqrt{2}-1\right)-\left(3-2\sqrt{2}\right)\right)+8-2\sqrt{2}\)
\(=-\left(\sqrt{-x^2-2x+3}-\sqrt{2}\right)^2-\left(x-\left(\sqrt{2}-1\right)\right)^2+8-2\sqrt{2}\le8-2\sqrt{2}\)
\(\Rightarrow E\le2\sqrt{2}-1\)
Dấu = xảy ra khi \(x=\sqrt{2}-1\)
\(C=3\sqrt{x}+2\sqrt{1-4x}\)
\(\Leftrightarrow2C=6\sqrt{x}+4\sqrt{1-4x}\)
\(=\left(-10x+6\sqrt{x}-\frac{9}{10}\right)+\left(-\frac{5}{2}\left(1-4x\right)+4\sqrt{1-4x}-\frac{8}{5}\right)+5\)
\(=-10\left(x-2.\sqrt{x}.\frac{3}{10}+\frac{9}{100}\right)-\frac{5}{2}\left(\sqrt{1-4x}-2.\left(1-4x\right).\frac{4}{5}+\frac{16}{25}\right)+5\)
\(=-10\left(\sqrt{x}-\frac{3}{10}\right)^2-\frac{5}{2}\left(\sqrt{1-4x}-\frac{4}{5}\right)^2+5\le5\)
\(\Rightarrow C\le\frac{5}{2}\)
Dấu = xảy ra khi \(x=\frac{9}{100}\)
a: ĐKXĐ: (x-1)(x-3)>=0
=>x>=3 hoặc x<=1
b: ĐKXĐ: (x-4)(x-3)>=0
=>x>=4 hoặc x<=3
c: ĐKXĐ: (x-5)(x-4)>=0
=>x>=5 hoặc x<=4
Ta có :
`-x_1^2+4x+12 = -(x_1^2 -4x-12) = -(x_1^2 -4x +4 -16) = -(x_1^2 -4x+4)+16 = -(x_1-2)^2+16 <= 16 AA x`
`-> \sqrt{-x_1^2+4x+12} <= \sqrt{16} =4 AA x`
Tương tự :
`-x_2^2 -2x+3 = -(x_2^2 -2x-3) = -(x_2^2 -2x+1-4) =-(x_2-1)^2+4 <= 4 AA x`
`-> \sqrt{-x_2^2 -2x+3} <= 2 AA x`
`-> C <= 4 -2 = 2`
Dấu `=` xảy ra : $x_1 =2$ và $x_2 =1$