Câu hỏi của Phoebe

Question

Tính giá trị biểu thức

S=$\dfrac{105}{abc+ab+a}$ +$\dfrac{b}{bc+b+1}$ +$\dfrac{a}{ab+a+105}$ biết : a, b, c $\in$ Z; abc=105 và bc+b+1$\ne$ 0

Nguyễn Huy Tú · Accepted Answer

Ta có: $S=\dfrac{105}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+105}$

$=\dfrac{abc}{a\left(bc+b+1\right)}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{a}{a\left(b+1+bc\right)}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{bc+b+1}$

$=\dfrac{bc+b+1}{bc+b+1}=1$

Vậy S = 1Câu trả lời: Ta có: $S=\dfrac{105}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+105}$

$=\dfrac{abc}{a\left(bc+b+1\right)}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{a}{a\left(b+1+bc\right)}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{bc+b+1}$

$=\dfrac{bc+b+1}{bc+b+1}=1$

Vậy S = 1

Hoang Hung Quan · Answer

Thay $abc=105$ ta có:

$S=\dfrac{abc}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$\Rightarrow S=\dfrac{abc}{a\left(bc+b+1\right)}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$\Rightarrow S=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{b+1+bc}$

$\Rightarrow S=\dfrac{bc+b+1}{bc+b+1}=1$

Vậy $S=1$Câu trả lời: Thay $abc=105$ ta có:

$S=\dfrac{abc}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$\Rightarrow S=\dfrac{abc}{a\left(bc+b+1\right)}+\dfrac{b}{bc+b+1}+\dfrac{a}{ab+a+abc}$

$\Rightarrow S=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{b+1+bc}$

$\Rightarrow S=\dfrac{bc+b+1}{bc+b+1}=1$

Vậy $S=1$