\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\)\(\frac{bc+ac+ab}{abc}=0\)

\(\Leftrightarrow\)\(bc+ac+ab=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}bc=-ab-ac\\ac=-ab-bc\\ab=-bc-ac\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a^2+2bc=a^2+bc-ab-ac=\left(a-b\right)\left(a-c\right)\\b^2+2ac=b^2+ac-ab-bc=\left(b-a\right)\left(b-c\right)\\c^2+2ab=c^2+ab-bc-ac=\left(c-a\right)\left(c-b\right)\end{cases}}\)

\(A=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ac+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(c-a\right)\left(c-b\right)}\)

= \(\frac{bc\left(b-c\right)+b-c+ac\left(c-a\right)+c-a+ab\left(a-b\right)+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{bc\left(b-c\right)+ca\left(c-a\right)-ab\left(b-c\right)-ab\left(c-a\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{\left(b-c\right)\left(bc-ab\right)+\left(c-a\right)\left(ca-ab\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{\left(a-c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

a) đáp án A=1

b) B=0

c) C=1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ac}{abc}=0\)

\(\Rightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)

\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự: \(b^2+2ac=\left(b-c\right)\left(b-a\right)\)

\(c^2+2ab=\left(a-c\right)\left(b-c\right)\)

\(B=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ca+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}-\frac{ca+1}{\left(a-b\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)\)

\(=\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-b\right)-\left(ca+1\right)\left(b-c\right)+\left(ab+1\right)\left(a-b\right)\)

\(=\left(b-c\right)\left(bc+1-ca-1\right)+\left(a-b\right)\left(ab+1-ca-1\right)\)

\(=\left(b-c\right)\left(bc-ca\right)+\left(a-b\right)\left(ab-ca\right)\)

\(=\left(b-c\right)c\left(b-a\right)+\left(a-b\right)a\left(b-c\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

Vậy B = 1

a,b,c khác nhau đôi một nghĩa là từng cặp số khác nhau ,là:

+a khác b

+b khác c

+c khác a

\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0\)

Suy ra: \(ab==-\left(bc+ac\right)=-bc-ac\)

    \(bc=-\left(ab+ac\right)=-ab-ac\)

\(ac=-\left(ab+bc\right)=-ab-bc\)

Nên \(a^2+2ab=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự,ta cũng có: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\)

                               \(c^2+2ab=\left(c-a\right)\left(c-b\right)\)

Vậy \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

những câu còn lại tương tự,bn tự làm nhé
 

ta có 1/a+1/b+1/c=0

=>bc+ac+ab/abc+0

=>bc+ac+ab=0

=>bc=-ac-ab

     ac=-bc-ab

     ab=-bc-ac

A=1/(a^2+bc-ac-ab)+1/(b^2+ac-bc-ab)+1/(c^2+ab-bc-ac)

=1/c(a-c)-b(a-c)+1/b(b-c)-a(b-c)+1/c(c-b)-a(c-b)

=1/(a-b)(a-c)+1/(b-a)(b-c)+1/(a-c)(c-b)

=b-c-a+c+a-b/(a-c)(a-b)(b-c)=0

('/': dấu gạch ngang ở giữa phân số)

https://hoc24.vn/hoi-dap/question/794424.html

Lời giải:

Xét tử :

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)

\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)

\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)

\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Xét mẫu (tương tự bên tử)

\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)

\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Do đó:

\(A=\frac{1}{1}=1\)

khó quá xin lỗi nha em  mới hok lớp 7

Câu này lớp 7 tớ có làm. Cũng như cái mà gọi là áp dụng t/c dãy tỉ số bằng nhau và tỉ lệ thức. mình tính ra dc a, b. c rồi.