Bài 7 : 

a, Với \(x\ge0;x\ne1\)

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

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

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

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

b, Ta có : \(x+\sqrt{x}+1=x+\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

mà 2 dương 

=> \(\frac{2}{x+\sqrt{x}+1}>0\)( đpcm ) 

Bài 4 : 

a, Với \(a>0\)

\(P=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

b, Ta có : \(a-\sqrt{a}-2=0\Leftrightarrow\sqrt{a}=2;\sqrt{a}=-1\left(voli\right)\Rightarrow a=4\)

còn 5 6 ạ

chọn a

\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)ĐK : \(x\ge0;x\ne1\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}-3}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}-3}{x+2\sqrt{x}-3}\)

\(=\frac{7\sqrt{x}-8-5x}{x+2\sqrt{x}-3}=\frac{-5x+7\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(a,\sqrt{27}-2\sqrt{42}+\sqrt{75}=3\sqrt{3}-2\sqrt{3}\sqrt{14}+5\sqrt{3}\)

\(=\left(8-2\sqrt{14}\right)\sqrt{3}\)

\(b,3\sqrt{20}-\sqrt{45}-\sqrt{550}=12\sqrt{5}-3\sqrt{5}-\sqrt{110}\sqrt{5}=\left(9-\sqrt{110}\right)\sqrt{5}\)

\(c,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+5\right)^2}\)

\(=|\sqrt{10}-3|-|\sqrt{10}+5|=\sqrt{10}-3-\sqrt{10}-5=-8\)

cho mk xin đề

câu hỏi đâu

đề đây ạaa

bài nào vậy làm j thấy