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

b, \(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}=\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}=\frac{\left(b+1\right)\left(2a-1\right)}{\left(2a-1\right)\left(b+1\right)3}=\frac{1}{3}\)

\(S=\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\\ =\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}\\ =\frac{\left(2a-1\right)\left(b+1\right)}{3\left(b+1\right)\left(2a-1\right)}\\=\frac{1}{3}\)

a)\(=\frac{ab+a-b-1}{ab-b+a-1}=1\)(Nhân phá ngoặc)

b)\(=\frac{2a+2ab-b-1}{6ab-3b+6a-3}\)(Nhân phá ngoặc)

\(=\frac{2ab+2a-b-1}{3\left(2ab+2a-b-1\right)}=\frac{1}{3}\)

\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)

\(=-2a+3b-4c+2a+3b+4c\)

\(=6b\)

b) Khi \(a=2012,b=-1,c=-2013\) ta có :

\(A=6b=6\cdot\left(-1\right)=-6\)

Vậy \(A=-6\) khi \(a=2012,b=-1,c=-2013\)

Giải:

a) \(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\) 

\(A=-2a+3b-4c+2a+3b+4c\) 

\(A=\left(-2a+2a\right)+\left(3b+3b\right)+\left(-4c+4c\right)\)

 \(A=0+2.3b+0\) 

\(A=6b\)

b) Ta thay: \(a=2012;b=-1;c=-2013\) 

Ta có:

\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\) 

\(A=\left(-2.2012+-3.1--4.2013\right)-\left(-2.2012--3.1--4.2013\right)\) 

\(A=\left(-2.2012-3.1+4.2013\right)-\left(-2.2012+3.1+4.2013\right)\)

\(A=-2.2012-3.1+4.2013+2.2012-3.1-4.2013\) 

\(A=\left(-2.2012+2.2012\right)+\left(-3.1-3.1\right)+\left(4.2013-4.2013\right)\) 

\(A=0+2.-3.1+0\) 

\(A=-6\)

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)

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

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

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

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