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

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

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

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

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

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

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

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

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

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

a) đáp án A=1

b) B=0

c) C=1

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)

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.

\(\text{Ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.\)

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

\(\Rightarrow BT\text{hức}=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

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

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

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

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

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

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

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

Lâu ko lm đổi dấu hơi thừa ra!! ko hiểu chỗ nào thì ib mk giải thích cho

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

=> \(\frac{ab+bc+ac}{abc}=0\)

=> \(ab+bc+ac=0\)

=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)

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

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

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

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

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

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

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

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

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

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

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

\(=\frac{\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

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

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

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

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

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

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

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

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

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

Cách I:(((dành cho nhũng ai biết HĐT a³ + b³ + c³ = [(a + b + c)(a² + b²+ c²-ab-bc-ca)+3abc]))) 
Ta có: 
bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³ 
=abc(1/a³ + 1/b³ + 1/c³) 
=abc[(1/a + 1/b + 1/c)(1/a² + 1/b²+ 1/c²-1/ab-1/bc-1/ca)+3/abc](áp dụng HĐt trên) 
=abc.3/(abc)=3 
Cách II: 
Từ giả thiết suy ra: 
(1/a +1/b)³=-1/c³ 
=>1/a³+1/b³+1/c³=-3.1/a.1/b(1/a+1/b)=3...
=>bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³ 
=abc(1/a³ + 1/b³ + 1/c³) 
=abc.3/(abc)=3

Mik ko biết có đúng ko??