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Ta có \(Q=\frac{a^2-ab+b^2}{a^2+ab+b^2}=\frac{3a^2-3ab+3b^2}{3a^2+3ab+b^2}=\frac{a^2+ab+b^2+2a^2-4ab+2b^2}{3a^2+3ab+3b^2}\) \(=\frac{1}{3}+\frac{2\left(a-b\right)^2}{3a^2+3ab+3b^2}\)
. Xét \(a^2+ab+b^2\) \(=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
. Suy ra \(\frac{1}{3}+\frac{2\left(a-b\right)^2}{3a^2+3ab+3b^2}\ge\frac{1}{3}\) => \(MinQ=\frac{1}{3}\) khi \(a=b\)
. \(Q=\frac{a^2-ab+b^2}{a^2+ab+b^2}=\frac{3a^2+3ab+3b^2-2a^2-4ab-2b^2}{a^2+ab+b^2}\) \(=3-\frac{2\left(a+b\right)^2}{a^2+ab+b^2}\le3\)
. Suy ra \(MaxQ=3\) khi \(a=-b\)
. Kết luận ^^
ta có \(4=2a^2+\frac{b^2}{4}+\frac{1}{a^2}=a^2+a^2+\frac{b^2}{4}+\frac{1}{a^2}\ge4\sqrt[4]{\frac{a^2.a^2.b^2}{4a^2}}\)
Vậy\(\sqrt[4]{\frac{a^2b^2}{4}}\le1\Leftrightarrow a^2b^2\le4\Leftrightarrow-2\le ab\le2\)
Vậy \(2007\le ab+2009\le2011\)
maximize=3 khi b=-a
minimize =1/3 khi a=b
rảnh thì làm cho h fai ngủ r` (:|
Bài 2:
Bài 1:
\(a^2+b^2+c^2=14\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=14\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=14\Rightarrow ab+bc+ac=-7\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=14^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=196-2.49=98\)
Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
ta có : 3-Q=\(\dfrac{2\left(a+b\right)^2}{a^2+ab+b^2}\)>=0
\(\Rightarrow\) Max Q=3
ta có : Q-\(\dfrac{1}{3}\)= \(\dfrac{2\left(a-b\right)^2}{3\left(a^2+ab+b^2\right)}\)>=0
\(\Rightarrow\)Min Q=\(\dfrac{-1}{3}\)
Hãy dùng phương pháp tập thể dục như của Hung nguyen nhé
Theo bài ra , ta có :
\(Q=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}=\dfrac{a^2+ab+b^2-2ab}{a^2+ab+b^2}=1-\dfrac{2ab}{a^2+ab+b^2}\)
Vì a,b đồng thời không bằng không ta chia cả tử và mẩu cho 2ab , ta được
\(\dfrac{2a}{a^2+ab+b^2}=\dfrac{1}{\dfrac{a^2}{2ab}+1+\dfrac{b^2}{2ab}}=\dfrac{1}{\dfrac{a}{2b}+1+\dfrac{b}{2a}}\)
Vì a,b khác 0 =) a/2b , b/2a khác 0
Áp dụng BĐT cô si cho 2 số a/2b , b/2a khác 0
\(\Rightarrow\dfrac{a}{2b}+\dfrac{b}{2a}\ge2\sqrt{\dfrac{a}{2b}.\dfrac{b}{2a}}\)
\(\Rightarrow\dfrac{a}{2b}+\dfrac{b}{2a}\ge2\sqrt{\dfrac{1}{2}}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{a}{2b}+1+\dfrac{b}{2a}\ge1+\dfrac{1}{4}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{a}{2b}+1+\dfrac{b}{2a}}\le\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}\)
\(\Leftrightarrow1-\dfrac{1}{\dfrac{a}{2b}+1+\dfrac{b}{2a}}\le\dfrac{1}{5}\)
\(\Rightarrow Max_Q=\dfrac{1}{5}\Leftrightarrow\dfrac{a}{2b}=\dfrac{b}{2a}\Leftrightarrow\dfrac{a}{2b}-\dfrac{b}{2a}=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-b\end{matrix}\right.\)
mà a và b là hai số khác 0 =) a = b
Vậy GTLN của Q là 1/5 khi và chỉ khi a = b