\(A=\dfrac{2}{\dfrac{a}{b}+\dfrac{4b}{a}}+\dfrac{1}{3\left(\dfrac{a}{b}\right)^2+2}\le\dfrac{3}{5}\)

Đặt \(\dfrac{a}{b}=y\)

\(A=\dfrac{2}{y+\dfrac{4}{y}}+\dfrac{1}{3y^2+2}\le\dfrac{3}{5}\)

\(A\Leftrightarrow\left(3y^2-5y+2\right)^2\ge0\)

Mashiro Shiina

cần 1 lời giải đáp cụ thể

trên face có đấy,lên đó mà tìm

\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)

\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)

\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)

$\frac{2x}{25-4b^2}:\frac{1}{5+2b}$

\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)

\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)

$\frac{{\left(2-a\right)}^2}{2ab}.\frac{b}{\left(2-a\right)}+\frac{1}{2}$

\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)

$\frac{2b+2}{2b-b^2}:\frac{b+1}{b}+\frac{2b+2}{3b-6}$

\(2a^2+5b^2+2ab=1\Leftrightarrow\left(a-b\right)^2+\left(a+2b\right)^2=1\)

Đặt \(P=\dfrac{a-b}{a+2b+2}\Rightarrow P\left(a+2b\right)+2P=a-b\)

\(\Rightarrow2P=\left(a-b\right)-P\left(a+2b\right)\)

\(\Rightarrow4P^2=\left[\left(a-b\right)-P\left(a+2b\right)\right]^2\le\left(P^2+1\right)\left[\left(a-b\right)^2+\left(a+2b\right)^2\right]=P^2+1\)

\(\Rightarrow3P^2\le1\Rightarrow-\dfrac{1}{\sqrt{3}}\le P\le\dfrac{1}{\sqrt{3}}\)

\(\dfrac{4}{a+b}-\dfrac{2a^2+3b^2}{2a^3+3b^3}-\dfrac{2b^2+3a^2}{2b^3+3a^3}=\dfrac{\left(a-b\right)^2.\left(12b^4+12ab^3-a^2b^2+12a^3b+12a^4\right)}{\left(a+b\right)\left(2a^3+3b^3\right)\left(2b^3+3a^3\right)}\ge0\)

PS: Còn cách dùng holder nữa mà lười quá

holder Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath

\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)

Tương tự và cộng lại;

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)