tau méc thầy hùng

ko bieets banj oi

a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)

= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)

= 1
@Nguyen Thi Ngoc Linh

Cứu mình với 9:00 sáng nay mình nộp bài rùi

Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)

Dấu = xảy ra \(\Leftrightarrow a=b=c\)

Mà đẳng thức trên xảy ra dấu =

\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)

Bài kia tí nghĩ nốt, khó v

Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)

Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)

\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)

\(=\frac{11}{16}:\frac{11}{32}\)

\(=\frac{11}{16}.\frac{32}{11}\)

\(=2\)