a) ĐKXĐ : \(a>0;a\ne1\)

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

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

\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)

\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)

b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)

\(\Rightarrow0< a< \frac{4}{25}\)

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

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

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

\(=\sqrt{x-2}-1.\frac{\left(\sqrt{x-2}-1\right)}{\sqrt{x}-\sqrt{3}}\)

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

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

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

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

\(=\sqrt{x-2}-1.\frac{\left(\sqrt{x-2}-1\right)}{\sqrt{x}-\sqrt{3}}\)

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

Good luck !!! Rất vui vì giúp đc bạn <3

trường hợp thứ  nhất:  \(X=1\)hay \(X=0\)

thì  \(X^2=X\)

trường hợp thứ  hai : \(X>1\)

thì  \(X^2>X\)

chúc học tốt

\(2.\left(x-4\right).\sqrt{x-2}+\left(x-2\right).\sqrt{x+1}+2x-6=0\)

\(\Leftrightarrow2.\sqrt{x-2}.x-8\sqrt{x-2}+\sqrt{x+1}.x-2\sqrt{x+1}+2x-6=0\)

Đặt x = u, ta có:

\(\Leftrightarrow2u\left(u^2+2\right)-8u+\sqrt{\left(u^2+2\right)+1}.\left(u^2+2\right)-2\sqrt{\left(u^2+2\right)+1}+2\left(u^2+2\right)-6=0\)

\(\Leftrightarrow\hept{\begin{cases}u=1\\u=-\frac{\sqrt{10}-2}{3}\\u=-\sqrt{2}-2\end{cases}}\Leftrightarrow x=3\)

=> x = 3

Không chắc nhé :v

ĐK \(x\ge2\)

Pt 

<=> \(2\left(x-4\right)\left(\sqrt{x-2}-1\right)+\left(x-2\right)\left(\sqrt{x+1}-2\right)+6x-18=0\)

<=> \(2\left(x-4\right).\frac{x-3}{\sqrt{x-2}+1}+\left(x-2\right).\frac{x-3}{\sqrt{x+1}+2}+6\left(x-3\right)=0\)

<=> \(\orbr{\begin{cases}x=3\\\frac{2\left(x-4\right)}{\sqrt{x-2}+1}+\frac{x-2}{\sqrt{x+1}+2}+6=0\left(2\right)\end{cases}}\)

Pt (2) \(VT=\frac{2\left(x-2\right)}{\sqrt{x-2}+1}+6-\frac{4}{\sqrt{x-2}+1}+\frac{x-2}{\sqrt{x+1}+2}>0\forall x\ge2\)

=> Pt (2) vô nghiệm

Vậy x=3

Xét hiệu \(x^2-x=x\left(x-1\right)\). Chú ý rằng x = 0 và x = 1 làm cho các thừa số x và x - 1 bằng 0.

Vì \(x\ge0\) nên ta xét các trường hợp:

*Nếu 0 < x < 1 thì \(x>0,x-1< 0\), do đó \(x^2-x< 0\)nên \(x^2< x\)

*Nếu x > 1 thì x và x - 1 đều dương, do đó  \(x^2-x>0\)nên \(x^2>x\)

*Còn nếu a = 0 hoặc a = 1 thì \(x^2=x\)

\(A=\frac{2}{2-\sqrt{3}}-2\sqrt{3}\)

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

A = 4

A = \(\frac{2}{2-\sqrt{3}}\)  - \(2\sqrt{3}\)

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

= 4 + \(2\sqrt{3}\) - \(2\sqrt{3}\)

=4

#mã mã#