c) Ta có: \(\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{\sqrt{6}}{\sqrt{3}}-3\cdot\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{3}\)

=0

Câu 2: 

a: \(\sqrt{9x-9}+1=7\)

\(\Leftrightarrow3\sqrt{x-1}=6\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: \(\sqrt{9x+27}-\dfrac{1}{4}\sqrt{16x+48}+\sqrt{x+3}=9\)

\(\Leftrightarrow\sqrt{x+3}=3\)

hay x=6

b: PTHĐGĐ là;

ax^2=2

=>ax^2-2=0

Δ=0^2-4*a*(-2)=8a

Để (P) cắt (d) tại hai điểm pb thì 8a>0

=>a>0

=>x=căn 2/a hoặc x=-căn 2/a

=>vecto OA=(căn 2/a;0); vecto OB=(-căn 2/a;0); vecto AB=(2*căn 2/a;2)

Theo đề, ta có: vecto OA*vecto OB=0 hoặc vecto OA*vecto AB=0 hoặc vecto OB*vecto AB=0

=>-2*căn 2/a+2=0 hoặc 2*căn 2/a+2=0

=>căn 2/a=1

=>a=2

\(C=\left(\dfrac{\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{\sqrt{x}}{x+\sqrt{x}}\right):\left(1-\dfrac{1}{\sqrt{x}}\right)\)

\(C=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left(1-\dfrac{1}{\sqrt{x}}\right)\)

\(C=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

\(C=\dfrac{x-1-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(C=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(C=\dfrac{1}{x-1}\)

a) \(\Leftrightarrow x^2=\sqrt{4}\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)

b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)

\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)

c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)

\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)

\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)

\(\Leftrightarrow\sqrt{2-x}=4\)

\(\Leftrightarrow2-x=16\)

hay x=-14

\(A=\dfrac{\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{3}\cdot3\sqrt{2}\)

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