cho a+b+c+d= 2020 và $\frac{1}{a+b+c}$+ $\frac{1}{b+c+d}$+ $\frac{1}{c+d+a}$+ $\frac{1}{d+a+b}$ = $\frac{1}{20}$
tính A= $\frac{a}{b+c+d}$+ $\frac{b}{c+d+a}$+ $\frac{c}{d+a+b}$+ $\frac{d}{a+b+c}$
cho a+b+c+d= 2020 và $\frac{1}{a+b+c}$+ $\frac{1}{b+c+d}$+ $\frac{1}{c+d+a}$+ $\frac{1}{d+a+b}$ = $\frac{1}{20}$
tính A= $\frac{a}{b+c+d}$+ $\frac{b}{c+d+a}$+ $\frac{c}{d+a+b}$+ $\frac{d}{a+b+c}$
Đáp án: $A=97$
Giải thích các bước giải:
Ta có:
$A=\dfrac{a}{b+c+d}+\dfrac{b}{c+d+a}+\dfrac{c}{d+a+b}+\dfrac{d}{a+b+c}$
$\to A=(\dfrac{a}{b+c+d}+1)+(\dfrac{b}{c+d+a}+1)+(\dfrac{c}{d+a+b}+1)+(\dfrac{d}{a+b+c}+1)-4$
$\to A=\dfrac{a+b+c+d}{b+c+d}+\dfrac{a+b+c+d}{c+d+a}+\dfrac{a+b+c+d}{d+a+b}+\dfrac{a+b+c+d}{a+b+c}-4$
$\to A=(a+b+c+d)(\dfrac{1}{b+c+d}+\dfrac{1}{c+d+a}+\dfrac{1}{d+a+b}+\dfrac{1}{a+b+c})-4$
$\to A =2020\cdot\dfrac{1}{20}-4$
$\to A=97$