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co nhieu cau tuong tu tren mang ban tu tm hieu nhe
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
Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{}\text{}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)
Ta thấy: \(\text{}\text{}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)
\(\text{}\text{}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)
Do đó: P \(\ge2+4+8=14\)
Vậy: P(min)=14 khi: \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)
Bài làm
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}=a+a+2b+b+\frac{4}{a}+\frac{9}{b}\)
\(=\left(a+2b\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)\)
\(\ge8+2\sqrt{a\times\frac{4}{a}}+2\sqrt{b\times\frac{9}{b}}\)( Cauchy )
\(=8+4+6=18\)
Đẳng thức xảy ra khi a = 2 ; b = 3
=> MinP = 18 <=> a = 2 ; b = 3
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}\)
\(\Leftrightarrow P=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)+a+2b\)
Áp dụng BĐT AM-GM ta có:
\(P\ge2.\sqrt{a.\frac{4}{a}}+2.\sqrt{b.\frac{9}{b}}+a+2b=2.2+2.3+a+2b\ge4+6+8=18\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{9}{b}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Vậy \(P_{min}=18\)\(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Nếu \(\left[{}\begin{matrix}a=0\Rightarrow b=0\Rightarrow b=2a\\b=0\Rightarrow a=0\Rightarrow b=2a\end{matrix}\right.\) trái với giả thiết \(\Rightarrow ab\ne0\)
\(2a^2+11ab-3b^2=0\Rightarrow2\left(\frac{a}{b}\right)^2+11\left(\frac{a}{b}\right)-3=0\)
Đặt \(\frac{a}{b}=x\ne0;\pm\frac{1}{2}\Rightarrow2x^2+11x-3=0\Rightarrow11x=3-2x^2\)
\(T=\frac{a-2b}{2a-b}+\frac{2a-3b}{2a+b}=\frac{\frac{a}{b}-2}{\frac{2a}{b}-1}+\frac{\frac{2a}{b}-3}{\frac{2a}{b}+1}=\frac{x-2}{2x-1}+\frac{2x-3}{2x+1}\)
\(T=\frac{\left(x-2\right)\left(2x+1\right)+\left(2x-3\right)\left(2x-1\right)}{4x^2-1}=\frac{6x^2-11x+1}{4x^2-1}=\frac{6x^2-\left(3-2x^2\right)+1}{4x^2-1}\)
\(T=\frac{8x^2-2}{4x^2-1}=2\)