\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

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

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)

a) x (x+1) (x-1) - (x-1) (x²+x+1)= x³ - x² + x² - x - x³ + 1³

                                           = 1- x

\(A=\left(ab+bc+ca\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).\)

\(A=\frac{1}{b}+\frac{1}{a}+\frac{ab}{c}+\frac{bc}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{ca}{b}+\frac{1}{a}-\frac{bc}{a}-\frac{ac}{b}-\frac{ab}{c}\)

\(A=2\cdot\frac{1}{b}+2\cdot\frac{1}{a}+2\cdot\frac{1}{c}\)

\(A=2.\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{c}\right)\)

Đặt;\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=m\Rightarrow mabc=ab+bc+ca\)

\(\Rightarrow m^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow m^2-2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

Thay vào A=\(mabc.m-abc.\left(m^2-2\left(\frac{a+b+c}{abc}\right)\right)=m^2abc-abcm^2+2\left(a+b+c\right)\)

\(=2a+2b+2c\)

\(A=\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(A=\left(ab+bc+ca\right).\frac{ab+bc+ca}{abc}-\frac{abc\left(a^2b^2+b^2c^2+c^2a^2\right)}{a^2b^2c^2}\)

\(A=\frac{\left(ab+bc+ca\right)^2}{abc}-\frac{a^2b^2+b^2c^2+c^2a^2}{abc}\)

\(A=\frac{2abc\left(a+b+c\right)}{abc}\)

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

Thay \(ab+bc+ca=1\) ta có:

\(1+a^2=ab+bc+ca+a^2=b\left(c+a\right)+a\left(c+a\right)=\left(c+a\right)\left(a+b\right)\)

Tương tự: \(1+b^2=\left(b+c\right)\left(a+b\right);\) \(1+c^2=\left(c+a\right)\left(b+c\right)\)

\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

\(\Rightarrow\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}=1\). Vậy biểu thức đó rút gọn lại bằng 1.