THỰC HIỆN PHÉP TÍNH
\(B=\frac{1}{a\left(a-b\right)\left(a-c\right)}+\frac{1}{b\left(b-a\right)\left(b-c\right)}+\frac{1}{c\left(c-a\right)\left(c-b\right)}\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có :\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
Tương tự : \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(MTC=\left(a-b\right)\left(b-c\right)\left(c-s\right)\left(a+b+c\right)\)
Kí hiệu biểu thức đã cho bởi \(Q\),ta có :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Thuwcj hiện phép tính:
a, A=\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
Thực hiện phép tính
C=\(\frac{bc}{\left(a-b\right).\left(a-c\right)}+\frac{ac}{\left(b-a\right).\left(b-c\right)}+\frac{ab}{\left(c-a\right).\left(c-b\right)}\)
A=\(\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\)\(\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
thực hiện phép tính
Lời giải :
\(A=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ac\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{-bc\left(b-c\right)-ac\left(c-a\right)-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử số :
\(TS=-b^2c+bc^2-ac^2+a^2c-a^2b+ab^2\)
\(=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(=\left(a-b\right)\left[-a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Khi đó \(A=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Thực hiện phép tính sau với a, b, c đôi một khác nhau
1) \(A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
Thực hiện phép tính :
a) \(\left(\frac{x-y}{x+y}+\frac{x+y}{x-y}\right).\left(\frac{x^2+y^2}{2xy}+1\right).\frac{xy}{x^2+y^2}\)
b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)
P/s : Giúp chauu với nhá mấy bacc kewtt :>>
\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Sr nha
Kq mik nhầm
Ko phải -2b đâu mà = 0
Oce :) Thks Moon :33
\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
Thực hiện phép tính
p/s: nhờ mng giúp e e cần gấp trước chiều nay e cám ơn nhiềuu
\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(b-a\right)}\)
\(=-\frac{-c\left(a^2-b^2\right)+ab\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-\frac{\left(a-b\right)\left[-c\left(a+b\right)+ab+c^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{\left(a-b\right)\left(-ac-bc+ab+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-\left(a-b\right)\left[-b\left(c-a\right)+c\left(c-a\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{\left(a-b\right)\left(c-a\right)\left(-b+c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(c-a\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Cho a,b,c khác 1 và a+b+c=3. Tính\(A=\frac{\left(a-1\right)^2}{\left(b-1\right)\left(c-1\right)}+\frac{\left(b-1\right)^2}{\left(c-1\right)\left(a-1\right)}+\frac{\left(c-1\right)^2}{\left(a-1\right)\left(b-1\right)}\)