Cho a, b, c doi mot khac nhau
CMR: \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}+\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
JUP VS
cho a,b,c khac nhau tung doi mot. Tinh
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{-a^2}{\left(a-b\right)\left(c-a\right)}+\frac{-b^2}{\left(b-c\right)\left(a-b\right)}+\frac{-c^2}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(-a^2\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-b^2\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-a^2b+ca^2-b^2c+ab^2-c^2a+bc^2}{-a^2b-c^2a+ca^2-b^2c+ab^2+bc^2}=1\)
Vậy \(P=1.\)
Cho a,b,c la 3 so doi mot khac nhau va \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\frac{\Leftrightarrow a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\) Nhân hai vế với \(\frac{1}{b-c}\)
Tương tự ta có:\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng (1),(2),(3) ta được đpcm
Cho a,b,c doi mot khac nhau va\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR: \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Tính các tổng :
a) A =\(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
b) B = \(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
Tính các tổng :
a) \(A=\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
b) \(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
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 + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)
\(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(b-a\right)\left(b-c\right)\left(c-a\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\left(\frac{b}{c-a}+\frac{c}{a-b}\right)\)
\(\Rightarrow\frac{a}{b-c}=-\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab-c^2+ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Tương tự:
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)};\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Cộng lại:
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
P/S:Đây nha ! Bài này lớp 7 chắc xem xong key cũng chắc là đang khó hiểu nhỉ ? Đưa giấy bút ra rồi nháp vài cái là hiểu ngay thôi !
Có người nhờ giải ấy @gunny :33
cho a,b,c khác nhau CMR:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{b-c}+\frac{2}{c-a}+\frac{2}{a-b}.\)
Câu hỏi của Bùi Minh Quân - Toán lớp 9 - Học toán với OnlineMath
Cho a,b,c khác nhau. Chứng minh rằng \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{c-a}+\frac{1}{a-b}\)
Tương tự:
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b}{\left(b-c\right)\left(b-a\right)}+\frac{b-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)
Và: \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c}{\left(c-a\right)\left(c-b\right)}+\frac{c-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
=> đpcm
Cho a,b,c chứng minh
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
Ta có
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\left(1\right)\)
Tương tự ta có
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\left(2\right)\)
\(\frac{a-b}{\left(c-b\right)\left(c-a\right)}=\frac{1}{b-c}+\frac{1}{c-a}\left(3\right)\)
Từ (1) (2) và (3) ta có
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(đpcm\right)\)
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{c-b}{\left(a-b\right)\left(c-a\right)}=\frac{\left(c-a\right)+\left(a-b\right)}{\left(a-b\right)\left(c-a\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Làm tương tự ta được:\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(ĐPCM\right)\)