Nếu \(a+b+c=0\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+0.\dfrac{2}{abc}\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
a. Đề bài sai, chắc chắn thiếu dữ kiện \(a+b+c=0\)
b.
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=\sqrt{\left(1+\dfrac{1}{99}-\dfrac{1}{100}\right)^2}\)
\(=1+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{9901}{9900}\)
Lời giải:
a.
Cho $a=1; b=5; c=6$ thì thấy đề sai.
b.
\(1+\frac{1}{99^2}+\frac{1}{100^2}=(1+\frac{1}{99})^2-\frac{2}{99}+\frac{1}{100^2}\)
\(=(\frac{100}{99})^2-\frac{2}{99}+\frac{1}{100^2}=(\frac{100}{99}-\frac{1}{100})^2\)
\(\Rightarrow \sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\frac{100}{99}-\frac{1}{100}\)
