Toán cho a/b+c + b/a+c + c/a+b =1 tính a^2/b+c + b^2/a+c + c^2/a+b 17/09/2021 By Alexandra cho a/b+c + b/a+c + c/a+b =1 tính a^2/b+c + b^2/a+c + c^2/a+b
gt: a/(b+c) + b/(c+a) + c/(a+b) = 1 A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)] = a[a/(b+c) + 1 – 1] + b[b/(c+a) + 1 – 1] + c[c/(a+b) + 1 – 1] = a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) – b + c.(a+b+c)/(a+b) – c = (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] – (a+b+c) = (a+b+c) – (a+b+c) = 0 Trả lời
gt: a/(b+c) + b/(c+a) + c/(a+b) = 1
A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)]
= a[a/(b+c) + 1 – 1] + b[b/(c+a) + 1 – 1] + c[c/(a+b) + 1 – 1]
= a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) – b + c.(a+b+c)/(a+b) – c
= (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] – (a+b+c)
= (a+b+c) – (a+b+c) = 0