danh dau minh nha

Trả lời:

- Cha mẹ có công sinh thành dưỡng dục, nuôi con nên người;

- Bạn bè là người gần gũi, giúp ta có sức mạnh tinh thần

- Thử thánh, thất bại là bài học của sự thành công

- Cô giáo là người mẹ hiền, nâng đỡ cho bao thế hệ học sinh vượt qua mọi chông gai trong cuộc sống

Các cao nhân cho xin 1 k nha!

d, \(\frac{3x}{x+2}=\frac{3\left(x+2\right)-6}{x+2}=3-\frac{6}{x+2}\)

\(\Rightarrow x+2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

e, \(C=\frac{A}{B}>0\Rightarrow\frac{3x}{x+2}.\frac{x+2}{x^2+2}=\frac{3x}{x^2+2}>0\)

\(\Rightarrow3x>0\Rightarrow x>0\)vì \(x^2+2>0\)

Kết hợp với đk vậy \(x>0;x\ne\pm2\)

f, vừa hỏi thầy, nên quay lại làm nốt :> 

f, Để \(\left|C\right|>C\Rightarrow C< 0\)vì \(\left|C\right|\ge0\)

\(\Rightarrow C=\frac{3x}{x^2+2}< 0\Rightarrow3x< 0\Leftrightarrow x< 0\)

\(A=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)(ĐK: \(x\ge0,x\ne4,x\ne9\))

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

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

\(A< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{2}\)(\(x\ge0,x\ne4,x\ne9\)) 

\(\Leftrightarrow\frac{2\sqrt{x}-4-\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-5< 0\)

\(\Leftrightarrow x< 25\).

Vậy \(0\le x< 25,x\ne4,x\ne9\)thì \(A< \frac{1}{2}\).

\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\inℤ\Leftrightarrow\frac{3}{\sqrt{x}+1}\inℤ\)

suy ra \(\sqrt{x}+1\inƯ\left(3\right)=\left\{1,3\right\}\)vì \(x\inℤ\)

\(\Rightarrow x\in\left\{0,4\right\}\)

Đối chiếu điều kiện ta chỉ có \(x=0\)thỏa mãn. 

vậy câu hỏi của bạn là gì

6, \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)

7, \(7+2\sqrt{10}=7+2\sqrt{5.2}=5+2\sqrt{5.2}+2=\left(\sqrt{5}+\sqrt{2}\right)^2\)

8, \(5-2\sqrt{6}=5-2\sqrt{2.3}=3-2\sqrt{2.3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

9, \(\sqrt{x^2-y^2}-x+y=\sqrt{\left(x-y\right)\left(x+y\right)}-\left(x-y\right)\)

\(=\sqrt{x-y}\left(\sqrt{x+y}-\sqrt{x-y}\right)\)

10, \(3x-2\sqrt{x}=\sqrt{x}\left(3\sqrt{x}-2\right)\)

1, \(\sqrt{xy}-x=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\)

2, \(x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2\)

3, \(x\sqrt{y}-y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

4, \(2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}=2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)\)

\(=\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)\)

5, \(\sqrt{35}-\sqrt{14}=\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)\)

a, Ta có : \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

Thay vào A ta được : \(A=\frac{\sqrt{2}-1-3}{\sqrt{2}-1+2}=\frac{\sqrt{2}-4}{\sqrt{2}+1}=\left(\sqrt{2}-4\right)\left(\sqrt{2}-1\right)=6-5\sqrt{2}\)

b, Với \(x\ge0;x\ne4;9\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)

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

c, Ta có : \(\frac{B}{A}< 1\Rightarrow\frac{\sqrt{x}-3}{3\sqrt{x}}< 1\Leftrightarrow\frac{\sqrt{x}-3}{3\sqrt{x}}-1< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3-3\sqrt{x}}{3\sqrt{x}}< 0\Leftrightarrow\frac{-2\sqrt{x}-3}{3\sqrt{x}}< 0\)

\(\Rightarrow-2\sqrt{x}-3< 0\Leftrightarrow-2\sqrt{x}< 3\Leftrightarrow4x< 9\Leftrightarrow x< \frac{9}{4}\)

Kết hợp với đk vậy 0 =< x < 9/4 

d, \(\frac{\sqrt{x}-3}{3\sqrt{x}};3=\frac{9\sqrt{x}}{3\sqrt{x}}\Rightarrow9\sqrt{x}>\sqrt{x}-3\Rightarrow\frac{B}{A}< 3\)

a) Thay x vào A ta được : \(A=\frac{\sqrt{3-2\sqrt{2}}-3}{\sqrt{3-2\sqrt{2}+2}}=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}-3}{\sqrt{\left(\sqrt{2}-1\right)^2}+2}=\frac{\sqrt{2}-1-3}{\sqrt{2}-1+2}=\frac{-4+\sqrt{2}}{1+\sqrt{2}}\)

\(=\frac{\left(-4+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\frac{-4+5\sqrt{2}-2}{-1}=6-5\sqrt{2}\)

b) \(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+\sqrt{5}x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+\sqrt{5}x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

c) \(\frac{B}{A}< 1\Leftrightarrow\frac{\frac{3\sqrt{x}}{\sqrt{x}+2}}{\frac{\sqrt{x}-3}{\sqrt{x}+2}}< 1\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}-3}< 1\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)(1)

Vì \(2\sqrt{x}+3>0\)nên \(\left(1\right)\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)

Kết hợp với ĐK => Với \(\hept{\begin{cases}0\le x< 9\\x\ne4\end{cases}}\)thì B/A < 1

d) mình đang kẹt ý d) bạn thông cảm ;-;

a, Với \(x\ge0;x\ne9\)

 \(N=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

\(=\left(\frac{-3}{\sqrt{x}+3}\right):\left(\frac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=-\frac{3}{\sqrt{x}+3}:\left(\frac{-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)=-\frac{3}{\sqrt{x}+3}.\left(-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)^2}\right)\)

\(=\frac{3}{\sqrt{x}-2}\)

b, Ta có : \(N< 0\Rightarrow\frac{3}{\sqrt{x}-2}< 0\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)

Kết hợp với đk vậy 0 =< x < 4 

c, Ta có : \(\sqrt{x}-2\ge-2\Rightarrow\frac{3}{\sqrt{x}-2}\le-\frac{3}{2}\)

Dấu ''='' xảy ra khi x = 0 

Vậy GTLN của N bằng -3/2 tại x = 0 

d, \(\frac{3}{\sqrt{x}-2}\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

e, Ta có : \(x=7-4\sqrt{3}=7-2.2\sqrt{3}=4-2.2\sqrt{3}+3=\left(2-\sqrt{3}\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

Thay vào N ta được : \(\frac{3}{\sqrt{x}-2}\Rightarrow\frac{3}{2-\sqrt{3}-2}=\frac{3}{-\sqrt{3}}=-\sqrt{3}\)

ĐK: \(x\ge-4\).

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

\(\Leftrightarrow2\sqrt{x+4+2\sqrt{x+4}+1}-\sqrt{x+4}=4\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x+4}+1\right)^2}-\sqrt{x+4}=4\)

\(\Leftrightarrow2\left(\sqrt{x+4}+1\right)-\sqrt{x+4}=4\)

\(\Leftrightarrow\sqrt{x+4}=2\)

\(\Leftrightarrow x=0\)(thỏa mãn) 

\(2\sqrt{x+5+2\sqrt{x+4}}-\sqrt{x+4}=4\)( ĐK : x >= -4 )

\(\Leftrightarrow2\sqrt{\left(\sqrt{x+4}+1\right)^2}-\sqrt{x+4}=4\)

\(\Leftrightarrow2\left|\sqrt{x+4}+1\right|-\sqrt{x+4}=4\)

\(\Leftrightarrow2\left(\sqrt{x+4}+1\right)-\sqrt{x+4}=4\left(\sqrt{x+4}+1>0\right)\)

\(\Leftrightarrow2\sqrt{x+4}+2-\sqrt{x+4}=4\)

\(\Leftrightarrow\sqrt{x+4}=2\Leftrightarrow x=0\left(tm\right)\)