Toán Lớp 7: cho các số a,b,c thỏa mãn a.b.c khác 0 và 1/a + 1/b + 1/c = a+b/c + b+c/a + c+a/b = 1/3 tính s=a+b+c=2020
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Toán Lớp 7: cho các số a,b,c thỏa mãn a.b.c khác 0 và 1/a + 1/b + 1/c = a+b/c + b+c/a + c+a/b = 1/3 tính s=a+b+c=2020
Comments ( 2 )
$\text{Ta có :}$
$\text{$\frac{a+ b}{c}$ + $\frac{b+c}{a}$ + $\frac{c+a}{b}$ }$
$\text{= $\frac{a}{c}$ + $\frac{a}{b}$ + $\frac{b}{c}$ + $\frac{b}{a}$ + $\frac{c}{a}$ + $\frac{c}{b}$ }$
$\text{= a . ( $\frac{1}{c}$ + $\frac{1}{b}$ + b . ( $\frac{1}{c}$ + $\frac{1}{a}$ ) + c . ( $\frac{1}{a}$ + $\frac{1}{b}$ )}$
$\text{Mà $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{1}{3}$ }$
$\text{=> $\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{1}{3}$ – $\frac{1}{a}$}$
$\text{=> $\frac{1}{c}$ + $\frac{1}{a}$ = $\frac{1}{3}$ – $\frac{1}{b}$ }$
$\text{=> $\frac{1}{a}$ + $\frac{1}{b}$ = $\frac{1}{3}$ – $\frac{1}{c}$ }$
$\text{và $\frac{a+b}{c}$ + $\frac{b+c}{a}$ + $\frac{c+a}{b}$ = $\frac{1}{3}$ }$
$\text{=> a . ( $\frac{1}{c}$ + $\frac{1}{b}$ ) + b . ( $\frac{1}{c}$ + $\frac{1}{a}$ ) + c . ( $\frac{1}{a}$ + $\frac{1}{b}$ ) = $\frac{1}{3}$ }$
$\text{=> a . ( $\frac{1}{3}$ – $\frac{1}{a}$ ) + b. ( $\frac{1}{3}$ – $\frac{1}{b}$ ) + c. ( $\frac{1}{3}$ – $\frac{1}{c}$ ) = $\frac{1}{3}$ }$
$\text{=> $\frac{a}{3}$ – 1 + $\frac{b}{3}$ – 1 + $\frac{c}{3}$ – 1 = $\frac{1}{3}$ }$
$\text{⇔ $\frac{a+b+c}{3}$ = $\frac{10}{3}$ ⇔ a + b + c = 10 }$
$\text{=> S = a + b + c + 2020 < 10 + 2020 = 2030 }$
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