Toán Lớp 7: Cho: $\frac{3}{b+a}$=$\frac{2}{a+c}$=$\frac{1}{c+b}$
Tính P= $\frac{a+b-2021.c}{a+b+2022.c}$
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Toán Lớp 7: Cho: $\frac{3}{b+a}$=$\frac{2}{a+c}$=$\frac{1}{c+b}$
Tính P= $\frac{a+b-2021.c}{a+b+2022.c}$
Giúp mình với ạ
Comments ( 1 )
#Kaito kid
Có $\frac{3}{b+a}$=$\frac{2}{a+c}$=$\frac{1}{c+b}$
$\frac{3+2+1}{b+a+a+c+c+b}$=$\frac{6}{2a+2b+2c}$=$\frac{2.3}{2(a+b+c)}$=$\frac{3}{a+b+c}$
⇒$\frac{3}{a+b}$=$\frac{3}{a+b+c}$
⇒a+b+c=a+b
⇒c=0
⇒P=$\frac{a+b-2021.0}{a+b+2022.0}$=$\frac{a+b}{a+b}$=1
Vậy P=1