làm tiếp nè:

\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)

*)Nếu \(\sqrt{x-1}\)>3<=>x-1>9<=>x>10 thì \(\sqrt{x-1}\)-2>0 \(\sqrt{x-1}\)-3>0

Ta có:|\(\sqrt{x-1}\)-2|+|\(\sqrt{x-1}\)-3|=\(\sqrt{x-1}\)-2+\(\sqrt{x-1}\)-3=2\(\sqrt{x-1}\)-5

*)Nếu 2<\(\sqrt{x-1}\)<3<=>4<x-1<9... làm tiếp đi bận mất rồi

A = \(\frac{8}{\sqrt{5}-1}\)  - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )                                    

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

= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1

= 3

B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )

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

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

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

= 1 +\(\sqrt{x}\)

#mã mã#

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

\(=\frac{-2\sqrt{x}+1+1+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)

\(=\frac{-2\sqrt{x}+2+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)

tick cho mình nha

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}\) có ĐKXĐ là x>=1

\(=\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)-6\sqrt{x-1}+9}\)

\(=\sqrt{\sqrt{x-1}^2-4\sqrt{x-1}+2^2}+\sqrt{\sqrt{x-1}^2-6\sqrt{x-1}+3^2}\)

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

\(=\left(\sqrt{x-1}-2\right)+\left(\sqrt{x-1}-3\right)=2\sqrt{x-1}-5\) với x>5

\(=-\left(\sqrt{x-1}-2\right)-\left(\sqrt{x-1}-3\right)=-2\sqrt{x-1}+5\) với x<5

thấy lạ nha , nãy giờ toàn thấy tên Chung Tình Là Tui , yêu tao ư kiếp sau nhé , yêu ông ư tui thà nhảy lầu còn hơn vậy ?

\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

Vây...

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

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

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

a: Đặt \(x^2-4=a\)

Pt sẽ là \(a=3\sqrt{xa}\)

\(\Rightarrow a^2=9xa\)

\(\Leftrightarrow a\left(a-9x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)

hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)

d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)

Pt sẽ là 2a+b=ab+2

=>(b-2)(1-a)=0

=>b=2 và 1-a

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)