a,Đk: \(x>0\)

Sau khi rút gọn được M=\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

Có \(x^2-x=0\) <=> \(x\left(x-1\right)=0\)=>x-1=0(vì x>0)

<=>x=1(t/m)

Thay x=1 vào b/thức M đã rút gọn có:

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

b, Có \(M=\frac{\sqrt{x}-2}{\sqrt{x}}=1-\frac{2}{\sqrt{x}}\)

Để M \(\in Z\) <=> \(\frac{2}{\sqrt{x}}\in Z\) => \(\frac{2}{\sqrt{x}}\in N^+\)

Với \(x\in N^+\)=> \(\left[{}\begin{matrix}\sqrt{x}\in N^+\\\sqrt{x}\notin N^+\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\frac{2}{\sqrt{x}}\in N^+\left(tm\right)\\\frac{2}{\sqrt{x}}\notin N^+\left(ktm\right)\end{matrix}\right.\)

=> \(\sqrt{x}\) thuộc ước tự nhiên của 2

<=> \(\sqrt{x}\in\left\{1,2\right\}\) <=> \(x\in\left\{1;4\right\}\)

Vậy để M\(\in Z< =>x\in\left\{1;4\right\}\)

mình cảm ơn bạn nhiều nha ^^

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)

\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)

\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )

Thay \(x=1\)vào M ta được:

\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)

c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)

Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)

\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)

Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )

Thử lại với \(x=4\)ta thấy M không là số tự nhiên

Vậy \(x\in\left\{0;16;36;144\right\}\)

em ko bieets hu hu

#)Giải :

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

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

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

\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)

#)Giải :

b) Để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với đkxđ => 0 < x < 9

Em không biết làm

a. x khac 1

b. 5-2√5 / 5

bạn làm chi tiết hơn đc không 

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

ĐK: \(x\ge0\)

=> \(x+\sqrt{x}\ge0\)

=> \(x+\sqrt{x}+1\ge1\)

=> \(\sqrt{x+\sqrt{x}+1}\ge1\)

=> \(-\sqrt{x+\sqrt{x}+1}\le1\)

Do đó: \(A\le4\)

Dấu "=" xảy ra khi x=0

\(B=\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+3}{1-\sqrt{x}}\left(ĐK:x\ge0;x\ne1\right)\)

\(=\frac{3x+6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(=\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

Dấu "=" xảy ra khi x=0

a)A= \(5-\sqrt{x+\sqrt{x}+1}\). ĐKXĐ: \(x\ge0\)

Ta luôn có: \(x+\sqrt{x}\ge0\) với \(x\ge0\)

\(\Rightarrow x+\sqrt{x}+1\ge1\)

\(\Rightarrow\sqrt{x+\sqrt{x}+1}\ge1\)

\(\Rightarrow-\sqrt{x+\sqrt{x}+1}\le-1\)

\(\Rightarrow5-\sqrt{x+\sqrt{x}+1}\le4\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy GTLN của A=4 khi x=0

b) B= \(\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{1-\sqrt{x}}\). ĐKXĐ: \(x\ge0; x\ne1\)

= \(\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

= \(\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) = \(\frac{x+2\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

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

Ta luôn có: \(\sqrt{x}+2\ge2\) với \(x\ge0; x\ne1\)

\(\Rightarrow\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\)

\(\Rightarrow1+\frac{1}{\sqrt{x}+2}\le\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy GTLN của B=\(\frac{3}{2}\) khi x=0