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Ta có: \(a^2+b^2=a+b\Leftrightarrow4a^2+4b^2=4a+4b\)
\(\Leftrightarrow4a^2-4a+4b^2-4b=0\Leftrightarrow\left(4a^2-4a+1\right)+\left(4b^2-4a+1\right)=2\)
\(\Leftrightarrow\left(2a-1\right)^2+\left(2b-1\right)^2=2\)
Áp dụng BĐT: \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
\(\Rightarrow\left(2a-1\right)^2+\left(2b-1\right)^2\ge\frac{\left(2a+2b-2\right)}{2}\)
\(\Rightarrow2\ge\frac{\left(2a+2b-2\right)^2}{2}\Leftrightarrow4\ge\left(2a+2b-2\right)^2\)
\(\Leftrightarrow1\ge a+b-1\Leftrightarrow4\ge a+b+2\)
Nhận thấy: \(S=\frac{a}{a+1}+\frac{b}{b+1}=\left(1-\frac{1}{a+1}\right)+\left(1-\frac{1}{b+1}\right)\)
\(=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\)
Ta áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+b+2}\Rightarrow2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\le2-\frac{4}{a+b+2}\)
Do \(a+b+2\le4\)(cmt) \(\Rightarrow\frac{4}{a+b+2}\ge1\Rightarrow2-\frac{4}{a+b+2}\le1\)
Từ đó: \(S=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\le2-\frac{4}{a+b+2}\le1\)
Suy ra \(Max\) \(S=1\).
Dấu "=" xảy ra khi \(a=b=1.\)
Ta CM BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
\(\Rightarrow a+b\ge\frac{\left(a+b\right)^2}{2}\)(do a2+b2=a+b)
\(\Rightarrow2\ge a+b\)
Ta có: \(S=\frac{a}{a+1}+\frac{b}{b+1}=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\)
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+1+b+1}\ge1\)
\(\Rightarrow S=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\le1\)
Dấu "=" xảy ra khi: a=b=1
Giải chi tiết dùm mình đi bạn, mình tick cho
Giải:
Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta xét các trường hợp:
Trường hợp \(1\): Nếu \(a+b+c=0\) thì:
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào \(P\) ta có:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{c}\right)\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)
Trường hợp \(2\): Nếu \(a=b=c\) thì:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2.2.2=8\)
Vậy \(P=-1\) hoặc \(P=8\)
ta có : a3+b3+c3-3abc=0
\(\Rightarrow\)(a+b)3+c3-3abc-3a2b-3ab2=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2+2ab-ac-bc)-3ab(a+b+c)=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2-ab-ac-bc)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\\\left(a+b+c\right)^2+a^2+b^2+c^2=0\Leftrightarrow a=b=c=0\left(bỏ\right)\end{matrix}\right.\)ta có P=(1+\(\dfrac{a}{b}\))(1+\(\dfrac{b}{c}\))(1+\(\dfrac{c}{a}\))
\(\Leftrightarrow\)p=\(\left(\dfrac{b+a}{b}\right)\left(\dfrac{c+b}{c}\right)\left(\dfrac{a+c}{a}\right)\)
\(\Leftrightarrow P=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)\)
\(\Leftrightarrow\)P=-1
a, P là snt > 3 => \(\left(p-1\right)\left(p+1\right)\)là tích 2 số chẵn liên tiếp ( p-1 >= 4 )
nên sẽ tồn tại 1 bội của 4 giả sử số đó là p+1
S uy ra \(p+1⋮4;p-1⋮2=>\left(p+1\right)\left(p-1\right)⋮8\)
Do P là snt lẻ > 3 => P sẽ có dạng 3k+1 hoặc 3k+2
rồi thay vồ => đpcm
\(x^2+xy-2019x-2020y-2021=x^2+xy+x-\left(2020x+2020y+2020\right)-1\)
\(=x\left(x+y+1\right)-2020\left(x+y+1\right)-1=\left(x-2020\right)\left(x+y+1\right)-1\)
làm tắt xíu :))
Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
\(a+b\ge a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\)
\(\Rightarrow2\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le1\)
Xét \(Q=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a\left(b+1\right)+b\left(a+1\right)}{\left(a+1\right)\left(b+1\right)}=\dfrac{a+b+2ab}{\left(a+1\right)\left(b+1\right)}\)
\(Q=\dfrac{a+b+ab+ab}{\left(a+1\right)\left(b+1\right)}\le\dfrac{a+b+ab+1}{\left(a+1\right)\left(b+1\right)}=\dfrac{\left(a+1\right)\left(b+1\right)}{\left(a+1\right)\left(b+1\right)}=1\)
\(\Rightarrow P\le2020+1^{2021}=2021\)
Dấu "=" xảy ra khi \(a=b=1\)