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 )

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

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

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

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

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

b, Ta có : \(M=\frac{4\sqrt{x}}{x+2\sqrt{x}+1}=\frac{8}{9}\Rightarrow36\sqrt{x}=8x+16\sqrt{x}+8\)

\(\Leftrightarrow8x-20\sqrt{x}+8=0\Leftrightarrow2x-5\sqrt{x}+2=0\)

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

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

\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)

\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)

Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)

Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)

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

Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

tích hộ mình nha

đề là  \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)đúng ko bạn

\(\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\left(\sqrt{2}-1\right)^2\right]\)

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

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

\(=2\left(6+1\right)\)

\(=14\)

\(b,\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

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

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

\(=\frac{6\sqrt{2}}{6}=\sqrt{2}\)

\(\hept{\begin{cases}\left(x+y\right)\left(x+y+z\right)=45\left(1\right)\\\left(y+z\right)\left(x+y+z\right)=63\left(2\right)\\\left(z+x\right)\left(x+y+z\right)=54\left(3\right)\end{cases}}\)

lấy (1)+(2)+(3) 

\(\left(x+y+z\right)\left(x+y+y+z+z+x\right)=162\)

\(\left(x+y+z\right)\left(2x+2y+2z\right)=162\)

\(\left(x+y+z\right)^2=81\)

\(x+y+z=9\)

thế vào pt (1)(2)(3) ta đc

\(\hept{\begin{cases}9\left(x+y\right)=45\\9\left(y+z\right)=63\\9\left(z+x\right)=54\end{cases}\hept{\begin{cases}x+y=5\\y+z=7\\z+x=6\end{cases}\hept{\begin{cases}x=5-y\left(4\right)\\z=7-y\left(5\right)\\z+x=6\left(6\right)\end{cases}}}}\)

thế(4);(5) vào (6)

\(5-y+7-y=6\)

\(y=3\)

\(x=5-3=2\)

\(z=7-3=4\)

thử lại lấy x+y+z \(3+2+4=9\left(TM\right)\)

góp ý nhỏ cho bài của bạn Như Quỳnh

Đoạn \(\left(x+y+z\right)^2=81\) ta phải có hai trường hợp là 

-9 và 9, trong cả hai trường hợp ta đều giải ra nghiệm thỏa mãn.

b. rõ ràng (0,0,0) là nghiệm của hệ

Xét x khác 0 dễ dàng ta chỉ ra được y và z khác 0

khi đó hệ \(\Leftrightarrow\hept{\begin{cases}6\left(\frac{1}{x}+\frac{1}{y}\right)=5\\12\left(\frac{1}{y}+\frac{1}{z}\right)=7\\4\left(\frac{1}{x}+\frac{1}{z}\right)=3\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=\frac{1}{2}\\\frac{1}{y}=\frac{1}{3}\\\frac{1}{z}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\\z=4\end{cases}}\) vậy hệ có hia nghiệm (0,0,0) và (2,3,4)

\(c.\frac{2x^2}{1+x^2}.\frac{2y^2}{1+y^2}.\frac{2z^2}{1+z^2}=xyz\Leftrightarrow\orbr{\begin{cases}xyz=0\\\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}=1\end{cases}}\)

\(xyz=0\Rightarrow x=y=z=0\) còn \(\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}=1\)

ta có \(1+x^2\ge2x\Leftrightarrow\frac{2x}{1+x^2}\le1\)

tương tự ta có : \(\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}\le1\)

Dấu = xảy ra khi x=y=z =1 

Vậy hệ có hai nghiệm (0,0,0) và (1,1,1)