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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(a>b>0\Rightarrow A=\frac{a+b}{a-b}>0\)
\(2a^2+2b^2=5ab\Rightarrow a^2+b^2=\frac{5ab}{2}\)
Ta có : \(E^2=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{\frac{5ab}{2}+2ab}{\frac{5ab}{2}-2ab}=\frac{\frac{9}{2}ab}{\frac{1}{2}ab}=\frac{\frac{9}{2}}{\frac{1}{2}}=9\)
\(E^2=9\Rightarrow E=3\)(vì E>0)
Vậy \(E=3\)
Có : \(2a^2+2b^2=5ab\Rightarrow\hept{\begin{cases}2a^2+2b^2-4ab=ab\\2a^2+2b^2+4ab=9ab\end{cases}}\Rightarrow\hept{\begin{cases}2\left(a-b\right)^2=ab\\2\left(a+b\right)^2=9ab\end{cases}}\Rightarrow\hept{\begin{cases}a-b=\sqrt{\frac{ab}{2}}\\a+b=\sqrt{\frac{9ab}{2}}\end{cases}}\)
\(\Rightarrow E=\frac{\sqrt{\frac{9ab}{2}}}{\sqrt{\frac{ab}{2}}}=\sqrt{\frac{\frac{9ab}{2}}{\frac{ab}{2}}}=\sqrt{\frac{9ab}{2}.\frac{2}{ab}}=\sqrt{9}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{9a^2-b^2}=\frac{3a^2+15ab-6b^2}{9a^2-b^2}\)\(=\frac{3a^2+3\left(3b^2-10a^2\right)-6b^2}{9a^2-b^2}=\frac{-3\left(9a^2-b^2\right)}{9a^2-b^2}=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: 2a2 + 2b2 = 5ab => 2(a2 + b2) = 5ab => a2 + b2 = \(\frac{5}{2}\)ab
\(A=\frac{2b}{a-b}+1=\frac{2b+a-b}{a-b}=\frac{a+b}{a-b}=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{\frac{5}{2}ab+2ab}{\frac{5}{2}ab-2ab}=\frac{\frac{9}{2}ab}{\frac{1}{2}ab}=9\)
Vậy A = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 5 nhé:
a) (a+1)2>=4a
<=>a2+2a+1>=4a
<=>a2-2a+1.>=0
<=>(a-1)2>=0 (luôn đúng)
vậy......
b) áp dụng bất dẳng thức cô si cho 2 số dương 1 và a ta có:
a+1>=\(2\sqrt{a}\)
tương tự ta có:
b+1>=\(2\sqrt{b}\)
c+1>=\(2\sqrt{c}\)
nhân vế với vế ta có:
(a+1)(b+1)(c+1)>=\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}\)
<=>(a+1)(b+1)(c+1)>=\(8\sqrt{abc}\)
<=>(a+)(b+1)(c+1)>=8 (vì abc=1)
vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án đúng : A
Dấu “=” xảy ra ⇔ 2 a − 1 3 − 2 a ≥ 0 ⇔ 1 2 ≤ a ≤ 3 2
Vậy GTNN của B là 2 khi 1 2 ≤ a ≤ 3 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(2a^2+2b^2=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow2a^2-ab-4ab+2b^2=0\)
\(\Leftrightarrow a\left(2a-b\right)-2b\left(2a-b\right)=0\)
\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2a-b=0\\a-2b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2a=b\\a=2b\end{cases}}}\)
Vì a>b>0 nên 2a>b
\(\Rightarrow a=2b\)
Thay vào P ta có
\(P=\frac{2.2b+b}{3.2b-b}=\frac{5b}{5b}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2a^2+\frac{1}{a^2}+\frac{b^2}{4}=4\Leftrightarrow\left(a^2+\frac{1}{a^2}-2\right)+\left(a^2+\frac{b^2}{4}-ab\right)=4-ab-2\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(a-\frac{b}{2}\right)^2=2-ab\)
\(VF=2-ab=\left(a-\frac{1}{a}\right)^2+\left(b-\frac{b}{2}\right)^2\ge0\)
Hay \(ab\le2\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=\frac{1}{a}\\b=\frac{b}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(a;b\right)=\left(1;\frac{1}{2}\right)\\\left(a;b\right)=\left(-1;-\frac{1}{2}\right)\end{cases}}\)
\(4a^2-5ab+b^2=0\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)
\(\Leftrightarrow4a-b=0\Rightarrow b=4a\)
\(\Rightarrow P=\frac{a.4a}{4a^2-\left(4a\right)^2}=\frac{4}{4-16}=-\frac{1}{3}\)