Viết rõ đề bài ra đc không ạ

đấy là phân số

Biến đổi được:  x = 2 ( a + b ) 3 ( a 3 − b 3 ) ; y = 9 ( a − b ) 2 4 ( a + b )

⇒ P = x . y = 2 ( a + b ) 3 ( a 3 − b 3 ) . 9 ( a − b ) 2 4 ( a + b ) = 3 ( a − b ) 2 ( a 2 + ab + b 2 )

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)

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

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

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

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

\(\left|a^2-3a+1\right|=1\)

=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)

ĐKXĐ: a<>2

=>a∈{0;1;3}

\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)

\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)

\(=2a^2-8a+1\)

Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)

Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)

Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)