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ĐKXĐ: ...
\(D=\left(\frac{2\sqrt{x}}{x\left(\sqrt{x}-1\right)+\sqrt{x}-1}-\frac{1}{\sqrt{x-1}}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\left(\frac{x+1}{x+\sqrt{x}+1}\right)\)
\(=\frac{\left(2\sqrt{x}-x-1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\frac{\left(x+1\right)}{\left(x+\sqrt{x}+1\right)}=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{1-\sqrt{x}}{x+\sqrt{x}+1}\)
b/ Do \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\) Để \(D>0\Leftrightarrow1-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 1\Rightarrow0\le x< 1\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
\(P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x-1}\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
mình chỉ rút gọn thôi , bạn tự so sánh nha 
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\)
=\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\)
=\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{x+1}\)