1. 

= -(13 + 3 căn7 ) / 2  +  -(7 + 3 căn7 ) / 2 

=  -7 + 3 căn7

Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

1.

đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)

có \(a^2+b^2=4\)

pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)

\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)

\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)

vì a,b>o nên \(a-b=\sqrt{2}\)

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

Bình phương 2 vế:

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

\(\Leftrightarrow\sqrt{4-x}=1\)

\(\Rightarrow x=3\)

Nếu đúng thì tích giùm mình cái nha!!!!!!!!!!!

2.ĐKXĐ D=R
Đặt \(a=\sqrt[3]{7-x},b=\sqrt[3]{x-5}\)
ta có: \(\hept{\begin{cases}a^3+b^3=2\\a^3-b^3=12-2x=2\left(6-x\right)\end{cases}}\)
Vậy ta có:

\(\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\Leftrightarrow\left(a-b\right)\left(2-\left(a+b\right)\left(a^2+ab+b^2\right)\right)=0\)
Th1: \(a-b=0\Leftrightarrow\sqrt[3]{7-x}=\sqrt[3]{x-5}\Leftrightarrow x=6\)
Th2: \(\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\a^3+b^3=12\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\\left(a+b\right)\left(a^2-ab+b^2\right)12\end{cases}}\)
Từ đó suy ra: 

\(\frac{a^2-ab+b^2}{a^2+ab+b^2}=6\Leftrightarrow5a^2-7ab+6b^2=0\)
nếu \(b=0\Leftrightarrow\sqrt[3]{x-5}=0\Leftrightarrow x=5\)thay vào phương trình ta thấy không thỏa mãn.
nếu \(b\ne0\Rightarrow5a^2-7ab+5b^2=0\Leftrightarrow5\left(\frac{a}{b}\right)^2-7\frac{a}{b}+5=0\)(1)
phương trình (1) vô nghiệm với ẩn \(\frac{a}{b}\). nên trường hợp này không xảy ra.
vậy phương trình có duy nhất nghiệm x = 6.

đkxđ: \(x\ge0;x\ne4\)

\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

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

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)

\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)

\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)

đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

=\(\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

Cám ơn bạn mã mã , để mình làm nốt nhé :

\(A=\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

Để \(A>2\Rightarrow\frac{4\sqrt{x}+8}{\sqrt{x}+7}>2\)

\(\Rightarrow\frac{4\sqrt{x}+8}{\sqrt{x}+7}-2>0\)

\(\Rightarrow\frac{4\sqrt{x}+8-2\sqrt{x}-14}{\sqrt{x}+7}>0\)

\(\Rightarrow\frac{2\sqrt{x}-6}{\sqrt{x}+7}>0\)

Vì \(\sqrt{x}>0\Rightarrow\sqrt{x}+7>0\)\(\Rightarrow A>0\Leftrightarrow2\sqrt{x}-6>0\)

\(\Rightarrow2\left(\sqrt{x}-3\right)>0\Rightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow\sqrt{x}>3\Rightarrow\sqrt{x}>\sqrt{9}\Rightarrow x>9\)

Vậy để \(A>2\Leftrightarrow x>9\)

f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)

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

\(=x-y\)

b)\(\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}\)

\(=\sqrt{7-4\sqrt{7}+4}-\sqrt{16+6\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{9+6\sqrt{7}+7}\)

\(=\sqrt{7}-2-\sqrt{\left(3+\sqrt{7}\right)^2}\)(vì \(\sqrt{7}>2\))

\(=\sqrt{7}-2-3-\sqrt{7}=-5\)

c)\(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{xy}}{y}\)

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