Ta có:

\(4a^2+b^2=5ab\Leftrightarrow4a^2+b^2-4ab-ab=0\)

\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\4a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\left(ktm\right)\\4a=b\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow4a=b\)

\(\Rightarrow\dfrac{5ab}{3a^2+2b^2}=\dfrac{5a.4a}{3a^2+2.\left(4a\right)^2}=\dfrac{20a^2}{3a^2+32a^2}\)

\(=\dfrac{20a^2}{35a^2}=\dfrac{4}{7}\)

\(4a^2+b^2=5ab\)

\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Rightarrow b=4a\left(do.a\ne b\right)\)

\(\dfrac{5ab}{3a^2+2b^2}=\dfrac{20a^2}{3a^2+32a^2}=\dfrac{4}{7}\)

x^5+x^4+1=(x^5+x^4+1)-(x^3+x^2+x)+(x^2+x+1)

                =x^3.(x^2+x+1) - x(x^2+x+1)+(x^2+x+1)

                =(x^2+x+1)(x^3-x+1)

\(2a^2+5ab-2b^2\)

\(=2a^2+ab+4ab-2b^2\)

\(=a\left(2a+b\right)+2b\left(2a+b\right)\)

\(=\left(2a+b\right)\left(a+2b\right)\)

\(2a^2+2b^2=5ab\)

<=>   \(2a^2+2b^2-5ab=0\)

<=>  \(2a^2-4ab-ab+2b^2=0\)

<=>   \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)

<=>  \(\left(2a-b\right)\left(a-2b\right)=0\)

<=>  \(\orbr{\begin{cases}2a-b=0\left(L\right)\\a-2b=0\end{cases}}\)

=>  \(a=2b\)

=>  \(A=\frac{a+2b}{2a-b}=\frac{2b+2b}{2.2b-b}=\frac{4b}{3b}=\frac{4}{3}\)

đây có phải trong violympic ko vậy

đây mik ghi bị thiếu đề

Từ \(2a^2+2b^2=5ab\),ta có :

\(2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(2a^2-4ab\right)-\left(ab-2b^2\right)=0\)

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

từ đó hoặc a-2b=0,hoặc 2a-b=0.Nhưng do b>a>0 nên suy ra b=2a

Vậy : \(\dfrac{a+b}{a-b}=\dfrac{a+2a}{a-2a}=\dfrac{3a}{-a}=-3\)

a: Ta có: \(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\)

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

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

\(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\)

\(=\frac{4a+2b}{ab\left(a^2+1\right)}-\frac{2}{a}=\frac{4a+2b-2b\left(a^2+1\right)}{ab\left(a^2+1\right)}\)

\(=\frac{4a-2a^2b}{ab\left(a^2+1\right)}=\frac{2a\left(2-ab\right)}{ab\cdot\left(a^2+1\right)}=\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}\)

Ta có: \(A=\left(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\right):\left(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\right)\)

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

b:

Sửa đề: b>a>0

\(4a^2+b^2=5ab\)

=>\(4a^2-5ab+b^2=0\)

=>\(4a^2-4ab-ab+b^2=0\)

=>(a-b)(4a-b)=0

TH1: a-b=0

=>a=b

mà a>b

nên Loại

TH2: 4a-b=0

=>b=4a(nhận)

\(A=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)

\(=\frac{4a}{\left(2-a\cdot4a\right)\left(2a-4a\right)}=\frac{4a}{\left(2-4a^2\right)\left(-2a\right)}\)

\(=\frac{4a}{-2a\cdot\left(-2\right)\left(2a^2-1\right)}=\frac{1}{2a^2-1}\)

bị nhầm xin lỗi

\(\Rightarrow\left(a+b\right)^2=\frac{9ab}{2};\left(a-b\right)^2=\frac{ab}{2}\)

Suy ra: \(\frac{2b}{a-b}+1=\frac{a+b}{a-b}=\frac{\frac{9ab}{2}}{\frac{ab}{2}}=9\)