\(Từ\) \(giả\) \(thiết\) : \(4a^2+b^2=\text{5}ab\)

\(\Leftrightarrow4a^2-4ab-ab+b^2\)

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

\(TH1:\) \(4a-b=0\) \((\) \(mẫu\) \(thuẫn\) \(với\) \(2a>b\) \()\)

\(TH2:\) \(a-b=0\)

\(\Rightarrow a=b\)

\(\Rightarrow A=\dfrac{a^2}{4a^2-a^2}\)

\(\Rightarrow A=\dfrac{1}{3}\)

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 \(P=\frac{1}{3}\)

ta có\(4a^2+b^2=5ab\)

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

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

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

thay (2a-b)² = ab vào P ta được

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

ta có: \(4a^2+b^2=5ab< =>4a^2-5ab+b^2=0< =>4a^2-4ab-ab+b^2=0< =>4a\left(a-b\right)-b\left(a-b\right)=0< =>\left(a-b\right)\left(4a-b\right)=0\)

do 2a>b>0=>4a>b>0=> 4a-b khác 0

=> a-b=0<=>a=b

P=\(\dfrac{ab}{4a^2-b^2}=\dfrac{ab}{\left(2a-b\right)\left(2a+b\right)}=\dfrac{ab}{\left(2a-a\right)\left(2a+a\right)}=\dfrac{a^2}{3a^2}=\dfrac{1}{3}\)

vậy............

chúc bạn hcoj tốt ^^

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 . . .