Question

Cho 3 số \(a,b,c\) thỏa mã điều kiện \(abc=105\) và \(bc+b+1\ne0\). Tính giá trị của biểu thức:

S=\(\frac{105}{abc+ab+a}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{a}{ab+a+105}\)

Answer 1

\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)

\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\left(abc=105\right)\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}\)

\(=1\)

Answer 2

\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)

\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)  \(\left(abc=105\right)\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}\)

\(=1\)