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Áp dụng Cauchy, ta có:
\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\)
\(\Rightarrow\frac{1}{a^4+b^2+2ab^2}\le\frac{1}{2a^2b+2ab^2}\)
Tượng tự:
\(\frac{1}{b^4+a^2+2a^2b}\le\frac{1}{2a^2b+2ab^2}\)
\(\Rightarrow A\le\frac{2}{2ab\left(a+b\right)}\)
Lại có: \(\frac{1}{a}+\frac{1}{b}=2\)\(\Leftrightarrow\frac{a+b}{ab}=2\Rightarrow a+b=2ab\)
\(\Rightarrow A\le\frac{2}{\left(a+b\right)^2}\)
Áp dụng Schwarzt: \(2=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge a+b\ge2\Rightarrow\left(a+b\right)^2\ge4\)
\(\Rightarrow A\le\frac{2}{4}=\frac{1}{2}\)
Dấu = xảy ra khi a=b=1
Áp dụng bđt cosi ta có :
A < = 1/2a^2b+2/ab^2 + 1/2ab^2+2a^2b
= 1/2ab . (1/a+b + 1/a+b) = 1/2ab . 2/a+b = 1/(a+b).(ab)
< = 1/\(\sqrt{ab}.2.ab\) = 1/2\(\sqrt{ab}^3\)
Có : 2 = 1/a + 1/b >= 2\(\sqrt{\frac{1}{ab}}\)
=> \(\sqrt{\frac{1}{ab}}\)< = 1
=> 1/ab < = 1
=> ab > =1
=> A < = 1/2.1 = 1/2
Dấu "=" xảy ra <=> a=b=1
Vậy GTLN của A = 1/2 <=> a=b=1
Tk mk nha
Lời giải:
\(\frac{1}{a}+\frac{1}{b}=1\Rightarrow a+b=ab\)
Khi đó:
\(A=\frac{(a^2-b^2)^2}{a^4b^4}+\frac{4}{ab}=\frac{(a-b)^2(a+b)^2}{(ab)^4}+\frac{4}{ab}\)
\(=\frac{(a-b)^2(ab)^2}{(ab)^4}+\frac{4}{ab}=\frac{(a-b)^2}{(ab)^2}+\frac{4ab}{(ab)^2}=\frac{(a-b)^2+4ab}{(ab)^2}=\frac{(a+b)^2}{(ab)^2}=\frac{(ab)^2}{(ab)^2}=1\)
Vậy.........
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Bạn xem lời giải ở đây nhé:
Câu hỏi của AgustD - Toán lớp 9 - Học toán với OnlineMath
\(\frac{1}{a}+\frac{1}{b}>=\frac{4}{a+b}\Rightarrow2>=\frac{4}{a+b}\Rightarrow a+b>=2\) (bđt cauchy schwarz adangj engel)
\(a^4+b^2>=2\sqrt{a^4b^2}=2a^2b;a^2+b^4>=2\sqrt{a^2b^4}>=2ab^2;\frac{1}{a}+\frac{1}{b}>=2\sqrt{\frac{1}{a}\cdot\frac{1}{b}}\Rightarrow2>=\frac{2}{\sqrt{ab}}\Rightarrow ab>=1\)(bđt cosi)
\(\Rightarrow\frac{1}{a^4+b^2+2ab^2}+\frac{1}{a^2+b^4+2a^2b}< =\frac{1}{2a^2b+2ab^2}+\frac{1}{2ab^2+2a^2b}=\frac{2}{2a^2b+2ab^2}=\frac{2}{2ab\left(a+b\right)}\)
\(=\frac{1}{ab\left(a+b\right)}< =\frac{1}{1\cdot2}=\frac{1}{2}\)
dấu = xảy ra khi a=b=1
1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)
<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)
<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)
Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)
Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
=\(\frac{5}{a+1}\)
Tương tự với x:y
\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)
\(a+\frac{1}{b}=1\)\(\Leftrightarrow\left(a+\frac{1}{b}\right)^2=1\)\(\Leftrightarrow a^2+\frac{1}{b^2}+\frac{2a}{b}=1\)\(\Leftrightarrow\frac{a}{b}=-1\)
\(a^2+\frac{1}{b^2}=3\)\(\Leftrightarrow\left(a^2+\frac{1}{b^2}\right)^2=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}+\frac{2.a^2}{b^2}=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}=7\)
\(N=\frac{a^4b^4+a^2b^2+1}{b^4}=a^4+\frac{a^2}{b^2}+\frac{1}{b^4}\)
\(\text{Thanks you verry much !!}\)