\(b)\left\{{}\begin{matrix}x-\dfrac{y}{2}=\dfrac{1}{2}\\\dfrac{x}{3}-2y=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-6y=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\2x-12y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=11\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x=1+1=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

\(c)\left\{{}\begin{matrix}5x-0,7y=1\\-10x+1,4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\5x-0,7y=1\end{matrix}\right.\) 

=> Hpt vô số nghiệm 

 Xét \(f\left(x\right)=VT=x^2+y^2+xy-3x-3y+3\)

\(=x^2+\left(y-3\right)x+y^2-3y+3\)

 Có \(\Delta=\left(y-3\right)^2-4\left(y^2-3y+3\right)\)

\(=y^2-6y+9-4y^2+12y-12\)

\(=-3y^2+6y-3\)

\(=-3\left(y-1\right)^2\le0\) với mọi \(y\inℝ\)

 Mà \(f\left(x\right)\) có hệ số cao nhất bằng \(1>0\) nên từ đây có \(VT=f\left(x\right)\ge0\)

 Dấu "=" xảy ra khi \(y=1\). Khi đó \(\Delta=0\) nên pt \(f\left(x\right)=0\) có nghiệm kép \(\Leftrightarrow\) \(x=\dfrac{-\left(y-3\right)}{2}=1\).

 Ta có đpcm.

a, Ta có: \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(SO_3+H_2O\rightarrow H_2SO_4\)

_____0,1____________0,1 (mol)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

______0,1_______0,2 (mol)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(\dfrac{4}{\sqrt{5}+\sqrt{3}}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}-\sqrt{2^2\cdot5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-2\sqrt{5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{2}-2\sqrt{5}\\ =2\left(\sqrt{5}-\sqrt{3}\right)-2\sqrt{5}\\ =2\sqrt{5}-2\sqrt{3}-2\sqrt{5}\\ =-2\sqrt{3}\)

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\\ =\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\ =2\sqrt{2}\)