Sửa đề: Chứng minh rằng \(\sqrt{\dfrac{1}{c^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\)
với \(a+b=c\) và \(a,b,c>0\)
Giải:
\(VT=\sqrt{\dfrac{1}{c^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}}\)
\(=\sqrt{\dfrac{a^2b^2+b^2c^2+a^2c^2}{a^2b^2c^2}}\)
\(=\dfrac{\sqrt{a^2b^2+c^2\left(a^2+b^2\right)}}{abc}\) (vì \(a,b,c>0\))
\(=\dfrac{\sqrt{a^2b^2+c^2\left[\left(a+b\right)^2-2ab\right]}}{abc}\)
\(=\dfrac{\sqrt{a^2b^2+c^2\left(c^2-2ab\right)}}{abc}\) (vì \(a+b=c\))
\(=\dfrac{\sqrt{a^2b^2-2abc^2+c^4}}{abc}\)
\(=\dfrac{\sqrt{\left(ab-c^2\right)^2}}{abc}\)
\(=\dfrac{c^2-ab}{abc}\) (vì \(c^2=a^2+2ab+b^2>ab\))
\(=\dfrac{c}{ab}-\dfrac{1}{c}\)
\(=\dfrac{a+b}{ab}-\dfrac{1}{c}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\left(\text{đ}pcm\right)\)
