giúp mình với m.n ơi.thanks trước nha!!!
Rút gọn phân thức sau:
Q=\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)+\(\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)+\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
K
Khách
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BA
8 tháng 11 2019
Ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)
tương tự
\(\frac{b^2-bc}{\left(a+b\right)\left(b+c\right)}=\frac{b}{a+b}-\frac{c}{b+c}\left(2\right)\)
\(\frac{c^2-ab}{\left(c+a\right)\left(b+c\right)}=\frac{c}{c+b}-\frac{a}{a+b}\left(3\right)\)
Cộng (1);(2) và (3) ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}=\frac{a}{a+c}-\frac{b}{a+b}+\frac{b}{a+b}-\frac{c}{b+c}+\frac{c}{c+b}-\frac{a}{a+b}=0 \)
19 tháng 2 2017
1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)
CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)
Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Thay vào biểu thwusc M ta được M=3abc (ĐPCM)
2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó
Nếu không thấy thì em có thể quy đồng lên mà rút gọn
8 tháng 7 2016
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 (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
MT
8 tháng 7 2016
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
8 tháng 7 2016
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 :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
AH
Akai Haruma
Giáo viên
4 tháng 7 2019
Lời giải:
\(\frac{1}{(b-c)(a^2+ac-b^2-bc)}+\frac{1}{(c-a)(b^2+ab-c^2-ac)}+\frac{1}{(a-b)(c^2+bc-a^2-ab)}\)
\(=\frac{1}{(b-c)[(a^2-b^2)+(ac-bc)]}+\frac{1}{(c-a)[(b^2-c^2)+(ab-ac)]}+\frac{1}{(a-b)[(c^2-a^2)+(bc-ab)]}\)
\(=\frac{1}{(b-c)(a-b)(a+b+c)}+\frac{1}{(c-a)(b-c)(b+c+a)}+\frac{1}{(a-b)(c-a)(c+a+b)}\)
\(=\frac{c-a}{(b-c)(a-b)(c-a)(a+b+c)}+\frac{a-b}{(a-b)(c-a)(b-c)(a+b+c)}+\frac{b-c}{(a-b)(c-a)(b-c)(a+b+c)}\)
\(=\frac{c-a+a-b+b-c}{(a-b)(b-c)(c-a)(a+b+c)}=0\)
19 tháng 12 2016
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
cảm ơn b nha ^^