Đáp án $+$ Giải thích Bài 1. Rút gọn,$a)~A=\frac1{(a-b)(a-c)}+\frac1{(b-c)(b-a)}+\frac1{(c-a)(c-b)}$,$b)~B=\frac1{a(a-b)(a-c)}+\frac1{b(b-c)(b-a)}+\frac1{c(c-a)(c-b)}$,$c)~C=\frac{bc}{(a-b)(a-c)}+\frac{ca}{(b-c)(b-a)}+\frac{ab}{(c-a)(c-b)}$

Question

Dương Quỳnh Anh · Accepted Answer

a) Ta có:

$A = \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)}$

Đổi dấu các mẫu thức để xuất hiện các nhân tử $(a-b)$, $(b-c)$, $(c-a)$:

$A = \frac{1}{(a-b)(a-c)} - \frac{1}{(b-c)(a-b)} + \frac{1}{(a-c)(b-c)}$

Quy đồng mẫu thức chung $(a-b)(b-c)(a-c)$:

$A = \frac{(b-c) - (a-c) + (a-b)}{(a-b)(b-c)(a-c)}$

$A = \frac{b - c - a + c + a - b}{(a-b)(b-c)(a-c)}$

$A = \frac{0}{(a-b)(b-c)(a-c)} = 0$

b) Ta có:

$B = \frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}$

Đổi dấu các mẫu thức:

$B = \frac{1}{a(a-b)(a-c)} - \frac{1}{b(b-c)(a-b)} + \frac{1}{c(a-c)(b-c)}$

Quy đồng mẫu thức chung $abc(a-b)(b-c)(a-c)$:

$B = \frac{bc(b-c) - ac(a-c) + ab(a-b)}{abc(a-b)(b-c)(a-c)}$

Biến đổi tử thức:

$bc(b-c) - ac(a-c) + ab(a-b) = b^2c - bc^2 - a^2c + ac^2 + a^2b - ab^2$

$= (a^2b - ab^2) - (a^2c - b^2c) + (ac^2 - bc^2)$

$= ab(a-b) - c(a^2-b^2) + c^2(a-b)$

$= ab(a-b) - c(a-b)(a+b) + c^2(a-b)$

$= (a-b)(ab - ac - bc + c^2)$

$= (a-b)[a(b-c) - c(b-c)]$

$= (a-b)(b-c)(a-c)$

Thay tử thức đã rút gọn vào $B$:

$B = \frac{(a-b)(b-c)(a-c)}{abc(a-b)(b-c)(a-c)}$

$B = \frac{1}{abc}$

c) Ta có:

$C = \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}$

Đổi dấu các mẫu thức:

$C = \frac{bc}{(a-b)(a-c)} - \frac{ca}{(b-c)(a-b)} + \frac{ab}{(a-c)(b-c)}$

Quy đồng mẫu thức chung $(a-b)(b-c)(a-c)$:

$C = \frac{bc(b-c) - ca(a-c) + ab(a-b)}{(a-b)(b-c)(a-c)}$

Tương tự câu b, tử thức biến đổi thành $(a-b)(b-c)(a-c)$.

Thay vào $C$:

$C = \frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(a-c)}$

$C = 1$

Timi · Answer

Xin lỗi, Timi không thể trả lời câu hỏi của bạn, hãy thử lại sau.

Huycindy · Answer

$a)\ A = \dfrac{1}{(a-b)(a-c)} + \dfrac{1}{(b-c)(b-a)} + \dfrac{1}{(c-a)(c-b)}$
$A = \dfrac{1}{(a-b)(a-c)} - \dfrac{1}{(a-b)(b-c)} + \dfrac{1}{(a-c)(b-c)}$
$A = \dfrac{b-c}{(a-b)(a-c)(b-c)} - \dfrac{a-c}{(a-b)(a-c)(b-c)} + \dfrac{a-b}{(a-b)(a-c)(b-c)}$
$A = \dfrac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}$
$A = \dfrac{0}{(a-b)(a-c)(b-c)}$
$A = 0$
$b)\ B = \dfrac{1}{a(a-b)(a-c)} + \dfrac{1}{b(b-c)(b-a)} + \dfrac{1}{c(c-a)(c-b)}$
$B = \dfrac{1}{a(a-b)(a-c)} - \dfrac{1}{b(a-b)(b-c)} + \dfrac{1}{c(a-c)(b-c)}$
$B = \dfrac{bc(b-c) - ac(a-c) + ab(a-b)}{abc(a-b)(a-c)(b-c)}$
$B = \dfrac{b^2c - bc^2 - a^2c + ac^2 + a^2b - ab^2}{abc(a-b)(a-c)(b-c)}$
$B = \dfrac{a^2(b-c) - a(b-c)(b+c) + bc(b-c)}{abc(a-b)(a-c)(b-c)}$
$B = \dfrac{(b-c)[a^2 - a(b+c) + bc]}{abc(a-b)(a-c)(b-c)}$
$B = \dfrac{(b-c)(a-b)(a-c)}{abc(a-b)(a-c)(b-c)}$
$B = \dfrac{1}{abc}$
$c)\ C = \dfrac{bc}{(a-b)(a-c)} + \dfrac{ca}{(b-c)(b-a)} + \dfrac{ab}{(c-a)(c-b)}$
$C = \dfrac{bc}{(a-b)(a-c)} - \dfrac{ca}{(a-b)(b-c)} + \dfrac{ab}{(a-c)(b-c)}$
$C = \dfrac{bc(b-c) - ca(a-c) + ab(a-b)}{(a-b)(a-c)(b-c)}$
$C = \dfrac{b^2c - bc^2 - a^2c + ac^2 + a^2b - ab^2}{(a-b)(a-c)(b-c)}$
$C = \dfrac{a^2(b-c) - a(b-c)(b+c) + bc(b-c)}{(a-b)(a-c)(b-c)}$
$C = \dfrac{(b-c)[a^2 - a(b+c) + bc]}{(a-b)(a-c)(b-c)}$
$C = \dfrac{(b-c)(a-b)(a-c)}{(a-b)(a-c)(b-c)}$
$C = 1$

Hurricane · Answer

{"userId":5447706,"userName":"Duy Hải"}

$a)$

$A=\dfrac{1}{(a-b)(a-c)}+\dfrac{1}{(b-c)(b-a)}+\dfrac{1}{(c-a)(c-b)}$

$A=\dfrac{1}{(a-b)(a-c)}+\dfrac{1}{(b-c)(-(a-b))}+\dfrac{1}{(-(a-c))(c-b)}$

$A=\dfrac{1}{(a-b)(a-c)}-\dfrac{1}{(b-c)(a-b)}-\dfrac{1}{(a-c)(c-b)}$

$A=\dfrac{b-c}{(a-b)(a-c)(b-c)}-\dfrac{a-c}{(a-b)(a-c)(b-c)}-\dfrac{a-b}{(a-b)(a-c)(b-c)}$

$A=\dfrac{b-c-a+c-a+b}{(a-b)(a-c)(b-c)}$

$A=\dfrac{0}{(a-b)(a-c)(b-c)}$

$A=0$

$b)$

$B=\dfrac{1}{a(a-b)(a-c)}+\dfrac{1}{b(b-c)(b-a)}+\dfrac{1}{c(c-a)(c-b)}$

$B=\dfrac{1}{a(a-b)(a-c)}-\dfrac{1}{b(a-b)(b-c)}+\dfrac{1}{c(c-a)(c-b)}$

$B=\dfrac{bc}{abc(a-b)(a-c)}-\dfrac{ac}{abc(a-b)(b-c)}+\dfrac{ab}{abc(c-a)(c-b)}$

$B=\dfrac{bc}{abc(a-b)(a-c)}+\dfrac{ac}{abc(b-c)(a-b)}+\dfrac{ab}{abc(c-a)(c-b)}$

$B=\dfrac{1}{abc}\left[\dfrac{bc}{(a-b)(a-c)}+\dfrac{ca}{(b-c)(b-a)}+\dfrac{ab}{(c-a)(c-b)}\right]$

$B=\dfrac{1}{abc}\left[\dfrac{bc}{(a-b)(a-c)}-\dfrac{ca}{(b-c)(a-b)}+\dfrac{ab}{(c-a)(c-b)}\right]$

$B=\dfrac{1}{abc}\cdot\dfrac{bc(b-c)-ca(a-c)+ab(a-b)}{(a-b)(b-c)(c-a)}$

$B=\dfrac{1}{abc}\cdot\dfrac{b^2c-bc^2-a^2c+ac^2+a^2b-ab^2}{(a-b)(b-c)(c-a)}$

$B=\dfrac{1}{abc}\cdot\dfrac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$

$B=\dfrac{1}{abc}\cdot\dfrac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}$

$B=\dfrac{1}{abc}$

$c)$

$C=\dfrac{bc}{(a-b)(a-c)}+\dfrac{ca}{(b-c)(b-a)}+\dfrac{ab}{(c-a)(c-b)}$

$C=\dfrac{bc}{(a-b)(a-c)}-\dfrac{ca}{(b-c)(a-b)}+\dfrac{ab}{(c-a)(c-b)}$

$C=\dfrac{bc(b-c)-ca(a-c)+ab(a-b)}{(a-b)(b-c)(c-a)}$

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

$C=\dfrac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$

$C=\dfrac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}$

$C=1$