Câu 1:

Áp dụng BĐT Cô-si:

\(A=\sqrt{\left(2-x\right)\left(2+x\right)}\le\frac{2-x+2+x}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow2-x=2+x\Leftrightarrow x=0\)

Câu 2:

\(B=\sqrt{-x^2+x+\frac{1}{4}}\)

\(B=\sqrt{-\left(x^2-x-\frac{1}{4}\right)}\)

\(B=\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{2}\right)}\)

\(B=\sqrt{-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\right]}\)

\(B=\sqrt{\frac{1}{2}-\left(x-\frac{1}{2}\right)^2}\le\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

\(x=2018-2\sqrt{2018}+1=\left(\sqrt{2018}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2018}-1\)

\(\Rightarrow P=\frac{\sqrt{2018}-1}{\sqrt{2018}-1+1}=\frac{\sqrt{2018}-1}{\sqrt{2018}}=\frac{2018-\sqrt{2018}}{2018}\)

1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)

\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)

\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)

Dấu '=' xảy ra khi x=0

2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)

Dấu '=' xảy ra khi x=0

3: \(A=-2x-3\sqrt{x}+2< =2\)

Dấu '=' xảy ra khi x=0

5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)

Dấu '=' xảy ra khi x=1

a) \(\sqrt{x-2}+\sqrt{16x-32}=10\)

\(\Rightarrow\sqrt{x-2}+4\sqrt{x-2}=10\)

\(\Rightarrow5\sqrt{x-2}=10\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

b) \(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)

ĐK \(x\ge\dfrac{1}{2}\)

\(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)

\(\left(\sqrt{x+\sqrt{2x-1}}\right)^2=\left(5\sqrt{2}\right)^2\)

\(\left|x+\sqrt{2x-1}\right|=50\)

\(\sqrt{2x-1}=50-x\)

\(\left(\sqrt{2x-1}\right)^2=\left(50-x\right)^2\)

\(\left|2x-1\right|=x^2-100x+2500\)

\(2x-1=x^2-100x+2500\)

\(x=41\)

câu 1 

ta có .....

lười viết Min - cốp xki nha

DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)

mà \(3x\ge-3\sqrt{5}\)

mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)

min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)