a) \(\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)

\(\Leftrightarrow ab+2b+1=ab+a+b+1\)

\(\Leftrightarrow b=a\)

Câu a sai đề, hình như pk là \(\frac{a}{b}=1\)

b) \(2\left(a+1\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)

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

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

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

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

\(\Leftrightarrow a^2-b^2=0\)

Hình như đề cx sai

a, b, c khác 0

Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:

 \(\frac{a+3b-c}{c}\)=\(\frac{-a+b+3c}{a}\) =\(\frac{c-b+3a}{b}\)=\(\frac{a+3b-c-a+b+3c+c-b+3a}{a+b+c}=\frac{3a+3b+3c}{a+b+c}=3\)

=> \(\frac{a+3b-c}{c}=3\Rightarrow\frac{a+3b}{c}-\frac{c}{c}=3\Rightarrow\frac{a+3b}{c}=4\)

\(\frac{-a+b+3c}{a}=3\Rightarrow-1+\frac{b+3c}{a}=3\Rightarrow\frac{b+3c}{a}=4\)

\(\frac{c-b+3a}{b}=3\Rightarrow\frac{c+3a}{b}-\frac{b}{b}=3\Rightarrow\frac{c+3a}{b}=4\)

=>  P =\(\left(3+\frac{a}{b}\right).\left(3+\frac{b}{c}\right).\left(3+\frac{c}{a}\right)=\frac{3b+a}{b}.\frac{3c+b}{c}.\frac{3a+c}{a}\)

= \(\frac{a+3b}{c}.\frac{b+3c}{a}.\frac{c+3a}{b}=4.4.4=64\)

a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)

Thay \(a=\frac{1}{2};x=-3\), ta có :

\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)

b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)

\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)

Thay \(a=-3;b=-4;c=2;d=3\), ta có :

\(B=\left(-3\right).\left(-4\right).2.3=72\)