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a) A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
=> A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)
=> A = \(\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)
\(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(b-a\right)\left(b-c\right)\left(c-a\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Lời giải chưa hay đâu bạn Trần Thị Kim Ngân.
Để ý một chút sẽ thấy \(A\) là một đa thức bậc 2 theo biến \(x\), nên ta gọi là \(A\left(x\right)\) cho đúng kiểu đa thức.
\(A\left(a\right)=1\) (nghĩa là thay \(x\) bằng \(a\) được kết quả là \(1\)).
Tương tự \(A\left(b\right)=A\left(c\right)=1\).
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Hừm, từ chỗ này về sau không biết bạn hiểu không.
Gọi \(f\left(x\right)=A\left(x\right)-1\) vẫn là một đa thức bậc 2, và \(f\left(a\right)=f\left(b\right)=f\left(c\right)=0\) tức là \(f\left(x\right)\) có 3 nghiệm \(x=1,x=b,x=c\).
Tuy nhiên, một đa thức bậc 2 thì chỉ có tối đa 2 nghiệm thôi, nếu nhiều hơn thì đa thức đó luôn bằng 0, nghĩa là \(f\left(x\right)=0\) với mọi \(x\).
Vậy \(A=1\).
Ta có:
\(A=\frac{\left(x-b\right)\left(x-c\right)\left(c-b\right)+\left(x-c\right)\left(x-a\right)\left(a-c\right)+\left(x-a\right)\left(x-b\right)\left(b-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{\left(x-b\right)\left(x-c\right)\left(c-b\right)+\left(x-c\right)\left(x-a\right)\left(a-c\right)-\left(x-a\right)\left(x-b\right)\left[\left(c-b\right)+\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{\left(x-b\right)\left(c-b\right)\left(x-c-x+a\right)+\left(x-a\right)\left(a-c\right)\left(x-c-x-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{\left(x-b\right)\left(c-b\right)\left(a-c\right)+\left(x-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-c\right)\left(c-b\right)\left(x-b-x+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
\(\frac{b+c+d}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}=\frac{\left(a+b+c+d-x\right)+\left(x-a\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}\)\(=\frac{\left(a+b+c+d-x\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}+\frac{1}{\left(b-a\right)\left(c-a\right)\left(d-a\right)}\)
Áp dụng hoán vị vòng \(b\rightarrow c\rightarrow d\rightarrow a\rightarrow b\) vào VT , ta được :
\(\left(a+b+c+d-x\right)\)[\(\frac{1}{\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(a-x\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)\left(b-d\right)\left(b-x\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)\left(c-d\right)\left(c-x\right)}\)\(+\frac{1}{\left(d-a\right)\left(d-b\right)\left(d-c\right)\left(d-x\right)}\).
Quy đồng mẫu thức và tính toán biểu thức trong [ ] ta được :
\(\frac{-1}{\left(x-a\right)\left(x-b\right)\left(x-c\right)\left(x-d\right)}\)
Vậy ...............
Ta có:\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(b-a\right)+\left(a-c\right)}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Chứng minh tương tự,ta được:
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)\left(đpcm\right)\)
\(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-c\right)\left(a-b\right)\left(b-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=0\)
\(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{-a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-ab+ac-bc+ab-ac+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=0\)