cho (d) : y = (2m - 3)x - 1. Tìm m sao cho khoảng cách từ gốc tọa độ đến (d) = 1/căn 5
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Ta có : x - y = 2 => x=2+y (1)
Mà 5x-3y=10 (2)
Thay (1) vào (2) ta dc : 5(2+y) - 3y =10
=> y = 0
=> x =0+2=2
\(5x-3y=10\)
\(\Leftrightarrow3\left(x-y\right)+2x=10\)
\(\Leftrightarrow6+2x=10\)
\(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{3\sqrt{3}+3}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{9+5\sqrt{3}}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}-\frac{2.\left(2\sqrt{3}+3\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{3\sqrt{3}+3+9+5\sqrt{3}-2\left(2\sqrt{3}+3\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{-2\left(2\sqrt{3}+3\right)+8\sqrt{3}+3+9}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{-2\left(2\sqrt{3}+3\right)+8\sqrt{3}+3+9}{9\sqrt{3}+15}\)
\(=\frac{4\sqrt{3}+6}{9\sqrt{3}+15}\)
\(=\frac{3-\sqrt{3}}{3}\)
b) \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}-2\right)\left(2\sqrt{5}+5\right)}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}-\frac{5\sqrt{5}+10}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}+\frac{20+9\sqrt{5}}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}-2\right)\left(2\sqrt{5}+5\right)-\left(5\sqrt{5}+10\right)+20-9\sqrt{5}}{20\sqrt{5}+45}\)
\(=\frac{5\sqrt{5}+10}{20\sqrt{5}+45}\)
\(=\frac{5\left(\sqrt{5}+2\right)}{5\left(4\sqrt{5}+9\right)}\)
\(=\frac{\sqrt{5}+2}{4\sqrt{5}+9}\)
\(=\sqrt{5}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\hept{\begin{cases}x-y=2\\5x-3y=10\end{cases}\Rightarrow\hept{\begin{cases}x-y=2\\2x+3x-3y=10\end{cases}}}\Rightarrow\hept{\begin{cases}x-y=2\\2x+3\left(x-y\right)=10\end{cases}\Rightarrow\hept{\begin{cases}x-y=2\\2x+6=10\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x-y=2\\2x=4\end{cases}\Rightarrow\hept{\begin{cases}x-y=2\\x=2\end{cases}\Rightarrow}\hept{\begin{cases}y=0\\x=2\end{cases}}}\)
Vậy y = 0 ; x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\frac{x^2-x+1}{x^2+x+1}=\frac{\frac{2}{3}x^2-\frac{4}{3}x+\frac{2}{3}}{x^2+x+1}+\frac{1}{3}=\frac{2}{3}\frac{\left(x-1\right)^2}{x^2+x+1}+\frac{1}{3}\ge\frac{1}{3}\)
\(\Rightarrow MIN\left(Q\right)=\frac{1}{3}\)Dấu "=" xảy ra khi x=1
\(Q=\frac{x^2-x+1}{x^2+x+1}=\frac{-2x^2-4x-2}{x^2+x+1}+3=-2\frac{\left(x+1\right)^2}{x^2+x+1}+3\ge3\)
\(\Rightarrow MAX\left(Q\right)=3\)Dấu "=" xảy ra khi x=-1