Question

a) Cho \(2x+5y=1\).Chứng minh rằng : \(x^2+y^2\ge\frac{1}{29}\).Dấu "=" xảy ra khi nào ?

b) Cho a> 0. Chứng minh rằng nếu : \(a-\frac{1}{a}=\sqrt{a}+\frac{1}{\sqrt{a}}\)thì ta có : \(a+\frac{1}{a}=3\)và \(a-\frac{1}{a}=\sqrt{5}\)

Answer 1

b/ \(a-\frac{1}{a}=\sqrt{a}+\frac{1}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}-\frac{1}{\sqrt{a}}=1\)

\(\Leftrightarrow a+\frac{1}{a}-2=1\)

\(\Leftrightarrow a+\frac{1}{a}=3\)

\(\Leftrightarrow a^2+\frac{1}{a^2}+2=9\)

\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2=5\)

\(\Leftrightarrow a-\frac{1}{a}=\sqrt{5}\)

Answer 2

a/ Ta có: \(x=\frac{1-5y}{2}\) thê vô ta được

\(x^2+y^2=y^2+\left(\frac{1-5y}{2}\right)^2=\frac{29y^2-10y+1}{4}\)

\(=\frac{1}{116}\left(29^2y^2-290y+29\right)=\frac{1}{116}\left[\left(29^2y^2-2.29y.5+25\right)+4\right]\)

\(=\frac{1}{116}\left[\left(29y-5\right)^2+4\right]\ge\frac{4}{116}=\frac{1}{29}\)