Toán Lớp 8: cho `a,b,c` thỏa mãn `ab + bc + ca = 6` CM: `a/{a^2 + 7} + b/{b^2 + 7} + c/{c^2 + b} = 12/{(a + b)(b + c)(c + a)}`

Question

Toán Lớp 8:  cho a,b,c thỏa mãn ab + bc + ca = 6 CM: a/{a^2 + 7} + b/{b^2 + 7} + c/{c^2 + b} = 12/{(a + b)(b + c)(c + a)}

thuminhthong · Answer

$  $   a/(a^2+6)   = a/(a^2+ab+bc+ca)   = a/(a(a+b)+c(a+b))   = a/((a+b)(a+c))   = (a(b+c))/((a+b)(a+c)(b+c))   = (ab+ac)/((a+b)(a+c)(b+c))   b/(b^2+6)   = b/(b^2 + ab + bc +ca)   = b/(b(a+b) +c(a+b))   =b/((a+b)(b+c))   =(b(a+c))/((a+b)(a+c)(b+c))   =(ab+bc)/((a+b)(a+c)(b+c))   c/(c^2+6)   =c/(c^2+ab+bc+ca)   =c/(c(b+c)+a(b+c))   =c/((b+c)(a+c))   =(c(a+b))/((a+b)(b+c)(c+a))   = (ac+bc)/((a+b)(b+c)(c+a))   a/(a^2+6)+b/(b^2+6)+c/(c^2+6)   = (ab+ac)/((a+b)(a+c)(b+c)) + (ab+bc)/((a+b)(a+c)(b+c)) + (ac+bc)/((a+b)(b+c)(c+a))   =(ab+ac+ab+bc+ac+bc)/((a+b)(b+c)(a+c))   =(2ab+2ac+2bc)/((a+b)(b+c)(a+c))   =(2.6)/((a+b)(b+c)(c+a))   =12/((a+b)(b+c)(c+a)) (= VP)   =>Đpcm

tuanh · Answer

Sửa đề: a/{a^2 + 6} + b/{b^2 + 6} + c/{c^2 + 6} = 12/{(a + b)(b + c)(c + a)}   ......   C&oacute;: a^2+6=a^2+ab+bc+ca   &hArr; a^2+6=(a^2+ab)+(bc+ca)   &hArr; a^2+6=a(a+b)+c(a+b)   &hArr; a^2+6 =(a+b)(a+c)   => a/{a^2 + 6} = a/{(a + b)(a + c)}   Chứng minh tương tự, ta được:   =>$ begin{cases}  dfrac{b}{b^2 + 6} =  dfrac{b}{(a + b)(b + c)}   dfrac{c}{c^2 + 6} =  dfrac{c}{(a + c)(b + c)} end{cases}$   VT = a/{a^2 + 6} + b/{b^2  + 6} + c/{c^2 + 6}          = a/{(a + b)(a + c)} +  $ dfrac{b}{(a + b)(b + c)}$ $ +  dfrac{c}{(a + c)(b + c)}$         = {a(b+c)+b(c+a)+c(a+b)}/{(a+b)(b+c)(c+a)}         = {ab+ac+ac+bc+ab+bc}/{(a+b)(b+c)(c+a)}           = {2(ab+bc+ac)}/{(a+b)(b+c)(c+a)}   m&agrave;: ab + bc + ca = 6   => = {2 xx 6}/{(a+b)(b+c)(c+a)}           = 12/{(a + b)(b + c)(c +a)}   => đpcm