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LD
0
1 tháng 2 2017
\(A=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{a}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac\left(bc+b+1\right)}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}=1\)
MQ
1
26 tháng 1 2017
cho mình xửa lại một chút nha:tính : A=\(\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}\)
CT
3
KT
23 tháng 12 2017
\(\frac{a}{ab+a+2}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{2c}{ac+2c+2}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{a\left(bc+b+1\right)}\)+ \(\frac{2abc}{ab\left(ac+2c+2\right)}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{abc+ab+a}\)+ \(\frac{2abc}{a^2bc+2abc+2ab}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{ab+a+2}\)+ \(\frac{2}{ab+a+2}\) (vì abc = 2 )
= \(\frac{ab+a+2}{ab+a+2}\)= 1
M
27 tháng 12 2020
M\(=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2bc}{b\left(ac+2c+2\right)}\)
M = \(\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{abc+2bc+2b}\)
M=\(\dfrac{1}{b+1+bc}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{2+2bc+2b}\)
M = \(\dfrac{1+b}{b+1+bc}+\dfrac{2bc}{2\left(1+bc+b\right)}\)
M = \(\dfrac{1+b}{b+1+bc}+\dfrac{bc}{b+1+bc}=\dfrac{1+b+bc}{b+1+bc}=1\)
DN
3
9 tháng 12 2018
Sửa đề:
\(\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)+\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}\)
\(=\frac{\left(a^2+b^2+c^2+ab+bc+ca\right)\left(a+b+c\right)}{a^2+b^2+c^2+2ab+2bc+2ca-\left(ab+bc+ca\right)}\)
\(=\frac{\left(a^2+b^2+c^2+ab+bc+ca\right)\left(a+b+c\right)}{a^2+b^2+c^2+ab+bc+ca}\)
\(=a+b+c\left(a^2+b^2+c^2+ab+bc+ca\ne0\right)\)
KT
2
AH
Akai Haruma
Giáo viên
22 tháng 2 2021
Lời giải:
\(\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{ab^2-ac^2-b^3+bc^2}=\frac{a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)}{a(b^2-c^2)-b(b^2-c^2)}\)
\(=\frac{(a^2-b^2)(b-c)-(b^2-c^2)(a-b)}{(a-b)(b^2-c^2)}=\frac{(a-b)(b-c)(a+b-b+c)}{(a-b)(b-c)(b+c)}=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}\)
\(=\frac{a-c}{b+c}\)
22 tháng 2 2021
Ta có: \(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)}{\left(b-c\right)\left(a-b\right)\left(b+c\right)}\)
\(=\dfrac{a-c}{b+c}\)