Ai giải giúp mình bài 1 với bài 4 trước đi

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

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

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

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

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

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

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

\(b,x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)

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

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

Thay \(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\) vào \(M=-\sqrt{2x}\) ta được:

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

\(M=-1-\sqrt{2}\)

Vậy ..............

ĐK: x > 0

a) Rút gọn M 

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

= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)

=> M \(\le\)1/3

=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn

Vậy max M = 1/3 tại x = 1

bn giải thíchcách làm câu b hôk mk vs mk ko hiểu

Giải thích lại nhé!

( Bạn có thể nói rõ là bạn không hiểu ở dòng nào?)

\(M=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

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

mà \(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\) ( theo cô - si )

=> \(\frac{1}{M}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)

=> \(M\le\frac{1}{3}\)

Dấu "=" xảy ra <=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\)<=> x = 1

Vậy GTLN của M là 1/3 đạt tại x = 1

$\left(\frac{1}{\sqrt{m}}+\frac{\sqrt{m}}{\sqrt{m}+1}\right):\frac{\sqrt{m}}{\sqrt{m}+m}$= \(\frac{\sqrt{m}+1+m}{\sqrt{m}\left(\sqrt{m}+1\right)}:\frac{\sqrt{m}}{\sqrt{m}+m}=\frac{m+\sqrt{m}+1}{\sqrt{m}+m}.\frac{\sqrt{m}+m}{\sqrt{m}}\)

\(=\frac{m+\sqrt{m}+1}{\sqrt{m}}=\sqrt{m}+1+\frac{1}{\sqrt{m}}\)

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

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

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(3\sqrt{x}+1\right)}\)

b) Để M>0 thì \(\frac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(3\sqrt{x}+1\right)}>0\)

mà \(\forall\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\), ta luôn có: \(\sqrt{x}\left(3\sqrt{x}+1\right)>0\)

nên \(\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)>0\)

mà \(\left(\sqrt{x}+1\right)^2>0\forall0< x\ne1\)

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\)

hay x>1(nhận)

Vậy: để M>0 thì x>1

a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)

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

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

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

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