Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(a+b+c\right)\left(ab+ac+bc\right)=\left(a+b+c\right)\left(ab+ac+bc+c^2-c^2\right)\)
\(=\left(a+b+c\right)\left(\left(a+c\right)\left(b+c\right)-c^2\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2\left(a+b\right)+c\left(a+c\right)\left(b+c\right)-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2a-c^2b+abc+c^2a+c^2b+c^3-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)+abc=\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018=2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Ta có:
\(A=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
\(A=\left(b^2c+abc\right)\left(c^2a+abc\right)\left(a^2b+abc\right)\)
\(A=bc\left(a+b\right)ac\left(b+c\right)ab\left(a+c\right)\)
\(A=\left(abc\right)^2\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=2018^2.0=0\)
Đề đúng phải là \(a^{2017}+b^{2017}=2.a^{1008}.b^{1008}\) nhé
Vì \(a^{2017}+b^{2017}=2.a^{1008}.b^{1008}\) nên \(\left(a^{2017}+b^{2017}\right)^2=4.a^{2016}.b^{2016}\)
Mà \(\left(a^{2017}+b^{2017}\right)^2\ge4.a^{2017}.b^{2017}\)
Suy ra \(4a^{2016}b^{2016}\ge4a^{2017}b^{2017}\)
<=> \(ab\le1\)
<=> \(1-ab\ge0\)
Suy ra P = 2018 - 2018ab = 2018(1 - ab) \(\ge0\)
\(a^{2017}+b^{2017}=2a^{2018}.b^{2018}\) với \(a,b\in R\)
nếu \(\orbr{\begin{cases}a=0\\b=0\end{cases}}\) thì \(P=2018>0\)
nếu \(\orbr{\begin{cases}a\ne0\\b\ne0\end{cases}}\) thì xảy ra 2 trường hợp như sau
\(TH1\)\(a,b\) trái dấu \(\Rightarrow P>0\)
\(TH2\) \(a,b\) cùng dấu
vì \(2.a^{2018}.b^{2018}>0\forall a,b\)
\(\Rightarrow a^{2017}+b^{2017}>0\) để 2 đẳng thức tồn tại dấu \("="\)
\(\Rightarrow a,b>0\) ( cùng dương)
có \(a^{2017}+b^{2017}=2a^{2018}.b^{2018}\)
\(\Leftrightarrow2=\frac{1}{a.b^{2018}}+\frac{1}{b.a^{2018}}\ge2\sqrt{\frac{1}{\left(a.b\right)^{2019}}}\)
\(\Rightarrow ab\le1\)
\(\Rightarrow2018-2018ab>2018-2018=0\)
dấu \("="\) xảy ra \(\Leftrightarrow a=b=1\)
vậy \(P\) luôn không âm
\(P=\left(b^2c+abc\right)\left(a^2b+abc\right)\left(c^2a+abc\right)\)
\(=bc\left(a+b\right)\cdot ab\left(c+a\right)\cdot ca\left(b+c\right)\)
\(=\left(abc\right)^2\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Lại có:
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)
\(\Leftrightarrow\left(a^2b+abc+a^2c\right)+\left(ab^2+b^2c+abc\right)+\left(bc^2+c^2a+abc\right)-abc=0\)
\(\Leftrightarrow a^2b+ca^2+ab^2+2abc+ac^2+b^2c+bc^2=0\)
\(\Leftrightarrow a^2\left(b+c\right)+a\left(b^2+2bc+c^2\right)+bc\left(b+c\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+a\left(b+c\right)^2+bc\left(b+c\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(a^2+ab+ca+bc\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)
\(\Rightarrow P=0\)
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: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)