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

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

=>a=b hoặc b=4a(loại)

=>P=b^2/3b^2=1/3

Ta có:

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

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

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

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

\(\Rightarrow4a=b\)

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

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

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

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

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

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

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

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

\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\Leftrightarrow\left(4a-b\right)\left(a-b\right)=0\)(1)

Ta thấy \(2a>b>0\left(gt\right)\) nên \(4a>b>0\Rightarrow4a-b>0\)

Từ đó để (1) xảy ra \(\Leftrightarrow a-b=0\Leftrightarrow a=b\) Thay vào P ta được :

\(P=\frac{ab}{4a^2-b^2}=\frac{a.a}{4a^2-a^2}=\frac{a^2}{3a^2}=\frac{1}{3}\)

Vậy \(P=\frac{1}{3}\)

P=1/3 nhé nhớ ko hở??^^

4a^2 + b^2=5ab 
<=>4a^2 + b^2 - 5ab=0 
<=>4a(a - b) - b(a - b)=0 
<=> (a -b )(4a - b)=0 
<=>a-b=0 ; a=b hoặc 4a - b=0 ; a=b/4(loại) 

đề lúc đầu sai :v 

ĐKXĐ : \(2a\ne b\)\(;\)\(2a\ne-b\)

\(4a^2+b^2=5ab\)\(\Leftrightarrow\)\(\left(a-b\right)\left(4a-b\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a-b=0\\4a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=b\\4a=b\end{cases}}}\)

+) Với \(a=b\)\(\Rightarrow\)\(M=\frac{ab}{4a^2-b^2}=\frac{a^2}{4a^2-a^2}=\frac{a^2}{3a^2}=\frac{1}{3}\)

+) Với \(4a=b\)\(\Rightarrow\)\(M=\frac{ab}{4a^2-b^2}=\frac{a.4a}{4a^2-16a^2}=\frac{4a^2}{-12a^2}=\frac{-1}{3}\)

... 

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

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

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

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

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

\(\left[\begin{array}{nghiempt}a-b=0\\4a-b=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}a=b\\4a=b\end{array}\right.\)

mà \(2a>b>0\)

\(\Rightarrow a=b\)

Thay a = b vào M, ta có:

\(M=\frac{b\times b}{4b^2-b^2}\)

\(=\frac{b^2}{3b^2}\)

\(=\frac{1}{3}\)

Vậy . . .

1/3 còn cách giải chờ mình 1 chút

Ta có: \(4a^2+b^2-5ab=0\Leftrightarrow4a^2-4ab+b^2-ab=0\Leftrightarrow4a\left(a-b\right)+b\left(b-a\right)=0\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)

nên \(a=b\) hoặc \(4a=b\)

Vì \(2a>b>0\Rightarrow\frac{2a}{b}>1\), ta lấy \(a=b\)

Thay \(a=b\) vào phân thức \(\frac{ab}{4a^2-4b^2}\), ta được:

\(A=\frac{1}{3}\)