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Ta có: \(\left(a^{100}+b^{100}\right)\cdot ab=a^{101}\cdot b+b^{101}\cdot a\)
\(\left(a^{101}+b^{101}\right)\cdot\left(a+b\right)=a^{102}+a^{101}\cdot b+b^{101}\cdot a+b^{102}\)
Do đó: \(\left(a^{101}+b^{101}\right)\left(a+b\right)-\left(a^{100}+b^{100}\right)\cdot ab\)
\(=a^{102}+b\cdot a^{101}+a\cdot b^{101}+b^{102}-a^{101}\cdot b-b^{101}\cdot a\)
\(=a^{102}+b^{102}\)
Kết hợp đề bài, ta có:
\(\left(a^{102}+b^{102}\right)\left(a+b\right)-\left(a^{102}+b^{102}\right)\cdot ab=a^{102}+b^{102}\)
\(\Leftrightarrow a+b-ab=1\)
\(\Leftrightarrow a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)+b\left(1-a\right)=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(1-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\1-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
Vậy: \(P=a^{2004}+b^{2004}=1^{2004}+1^{2004}=2\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel,ta có:
\(A=\frac{a^2}{a+1}+\frac{b^2}{b+1}\ge\frac{\left(a+b\right)^2}{a+b+2}=\frac{1}{1+2}=\frac{1}{3}^{\left(đpcm\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=1\\\frac{a}{a+1}=\frac{b}{b+1}\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=1\\ab+a=ab+b\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)
Vậy ...
\(P=2a+3b+\frac{1}{a}+\frac{4}{b}=a+2b+\left(a+\frac{1}{a}\right)+\left(b+\frac{4}{b}\right)\)
\(\ge5+2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{4}{b}}=5+2+4=11\)
Dấu "=" xảy ra <=> \(a=1;\)\(b=2\)
Vậy MIN P = 11 Khi a = 1; b = 2
Bài này là BĐT cosi
\(P=2a+3b+\frac{1}{a}+\frac{4}{b}\)
\(P=a+2b+\left(a+\frac{1}{a}\right)+\left(b+\frac{4}{b}\right)\)
\(P\ge5+2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{4}{b}}=5+2+4=11\)
Dấu "=" xảy ra khi a = 1/a <=> a = 1 ; b = 4/b <=> b = 2
Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)
\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)
\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Từ \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}=2\left(a^{101}+b^{101}\right)\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}-2\left(a^{101}+b^{101}\right)=0\)
\(\Rightarrow\left(a^{102}-2a^{101}+a^{100}\right)+\left(b^{102}-2b^{101}+b^{100}\right)=0\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2=0\left(1\right)\)
Vif \(\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2\ge0\forall a\\\left(b^{51}-b^{50}\right)^2\ge0\forall b\end{cases}}\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2\ge0\forall a,b\left(2\right)\)
Tứ (1) và (2) :
\(\Rightarrow\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2=0\\\left(b^{51}-b^{50}\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}-a^{50}=0\\b^{51}-b^{50}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}=a^{50}\\b^{51}=b^{50}\end{cases}}\)
Vì a,b là các số thực dương nên \(a=b=1\)
\(\Rightarrow P=a^{2007}+b^{2007}=1^{2007}+1^{2007}=1+1=2\)
Vậy \(P=2\)
co the la 0 ma
\(\left(a^{101}+b^{101}\right)\left(a+b\right)-ab\left(a^{100}+b^{100}\right)=a^{102}+a^{102}\Rightarrow\left(a^{102}+b^{102}\right)\left(a+b-ab\right)=a^{102}+b^{102}\Rightarrow a+b-ab=1\left(v\text{ì}:a,b>0\right)\Leftrightarrow a-ab+b-1=0\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=\left(a-1\right)\left(1-b\right)=0\Leftrightarrow\left[{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)
\(+,a=1\Rightarrow b=1\Rightarrow P=2\)
\(+,b=1\Rightarrow a=1\Rightarrow P=2\)
Vậy:P=2