\(a,ĐKXĐ:\hept{\begin{cases}a\ge0,\sqrt{a}\ne0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}}\)

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

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

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

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

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

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

c, bn thay vào rồi tính nha

ta có: \(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=1^2\)

\(\Leftrightarrow2x^2y^2+x^2+y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=0\)

\(\Leftrightarrow\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=0\)

\(\Leftrightarrow x\sqrt{1+y^2}+y\sqrt{1+x^2}=0\left(đpcm\right)\)

Đkxđ : với mọi x

pt <=> \(\sqrt{\left(2x-1\right)^2}\)= x+2

<=> 2x-1 = x+2

<=> 2x-x = 2+1

<=> x=3 (tm)

Vậy pt có tập nghiệm S = {3}

k mk nha

4x^2-4x+1=x^2+4x-4

4x^2-x^2-4x-4x+1+4=0

3x^2-8x+5=0

3x^2-3x-5x+5=0

3x(x-1)-5(x-1)=0

(x-1)(3x-5)=0

=>x=1 và x=5/3