\(A=\frac{3x}{4}+\frac{x}{4}+\frac{1}{x}\ge\frac{3x}{4}+2\sqrt{\frac{x}{4x}}\ge\frac{3.2}{4}+1=\frac{5}{2}\)

\(A_{min}=\frac{5}{2}\) khi \(x=2\)

\(B=\frac{24x}{25}+\frac{x}{25}+\frac{1}{x}\ge\frac{24x}{25}+2\sqrt{\frac{x}{25x}}\ge\frac{24.5}{25}+\frac{2}{5}=\frac{26}{5}\)

\(B_{min}=\frac{26}{5}\) khi \(x=5\)

Câu C bạn coi lại đề, nếu đúng thế này thì ko tồn tại min

a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)

đk: x >/ 0

(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)

Kl: \(x=\dfrac{392}{169}\)

b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)

đk: x >/ 5

(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)

Kl: x=9

c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)

Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)

(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)

Kl: x=-6

d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)

Đk: \(x\ge\dfrac{4}{5}\)

(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)

Kl: x=12

Minh bi nham dau bai, chi co 1 thua so \(\dfrac{2}{x}\) thoi nhe!

@Phùng Khánh Linh

@Aki Tsuki

@Nhã Doanh

@Akai Haruma

@Nguyễn Khang

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}\)

\(=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}=\sqrt{2}.\sqrt{2^2-2}=2\)

\(y=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{\frac{2}{3}(9\sqrt{2}-6\sqrt{3}+3\sqrt{5})}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{2}{3}\)

Do đó:

\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)

ĐKXĐ:...

\(M=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)

\(N=\frac{x\sqrt{x}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)

Để \(M=N\Leftrightarrow x-1=2\sqrt{x}+1\)

\(\Leftrightarrow x-2\sqrt{x}-2=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\left(\sqrt{3}+1\right)^2=4+2\sqrt{3}\)

@Hắc Hường

x=1

y=2

z=-3

==>x+y-z=1+2+3=6