Rút gọn phân thức sau\(\frac{a^2+b^2-c^2+2ab}{a^2+b^2+c^2+2ac}\)
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NC
1
8 tháng 12 2018
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)
\(=\frac{a+b-c}{a+c-b}\left(a+b+c\ne0\right)\)
25 tháng 11 2016
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
HN
25 tháng 11 2016
a) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
TT
2
12 tháng 10 2018
Ta có :
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
12 tháng 10 2018
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a+b\right)^2-2ab-c^2+2ab}{\left(a+c\right)^2-2ac-b^2+2ac}.\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)
\(=\frac{a+b-c}{a+c-b}\)
5 tháng 11 2016
a) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
9 tháng 2 2020
a, Gợi ý nà :3
a^2 + b^2 - c^2 +2ab = (a^2 + b^2 + 2ab) -c^2 = (a+b)^2 - c^2 = (a + b - c)(a + b + c)
a^2 - b^2 + c^2 + 2ac = (a + c)^2 - b^2 = (a + b + c)(a - b + c)
b. Gợi ý tiếp luôn nà :3
a^3 + b^3 + c^3 - 3abc
= (a^3 + b^3 +3a^2 x b + 3ab^2) - 3ab(a+b) -3abc + c^3
= (a+b)^3 + c^3 - 3ab(a+b+c)
= (a + b+ c)[(a+b)^2 - c(a+b) +c^2] - 3ab(a+b+c)
=(a+b+c)(a^2 + b^2 + c^2 -ac -bc + 2ab -3ab)
=(a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca)
Rồi cứ thế rút gọn...
Học tốt nha bạn :3
S
9 tháng 2 2020
\(\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\frac{a+b-c}{a-b+c}\)
\(\text{nhận xét: ta có hằng đẳng thức:}\)
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
đó đến đây bạn làm tiếp
LM
19 tháng 7 2017
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
LM
19 tháng 7 2017
b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
NV
3 tháng 1 2016
Câu này lớp 7 tớ có làm. Cũng như cái mà gọi là áp dụng t/c dãy tỉ số bằng nhau và tỉ lệ thức. mình tính ra dc a, b. c rồi.
19 tháng 1 2021
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=> \(\frac{ab+bc+ac}{abc}=0\)
=> \(ab+bc+ac=0\)
=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(=\frac{bc}{a^2-ab-ac+bc}+\frac{ca}{b^2-ab-bc+ac}+\frac{ab}{c^2-ac-bc+ab}\)
\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ca}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}-\frac{ca}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca^2-c^2a}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2-ca^2+c^2a+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(c^2a-bc^2\right)-\left(ca^2-b^2c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
19 tháng 1 2021
b) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-ab-bc+ac}+\frac{c^2}{c^2-bc-ac+ab}\)
\(=\frac{a^2}{a\left(a-b\right)-c\left(a-b\right)}+\frac{b^2}{b\left(b-a\right)-c\left(b-a\right)}+\frac{c^2}{c\left(c-b\right)-a\left(c-b\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2a-b^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+c+b\right)}\)
\(=\frac{a+b-c}{a+c-b}\)
Bạn sai đề nên mik sửa và làm luôn nha
\(a^2+b^2-c^2+2ab\)
______________________
\(a^2+b^2+c^2+2ac\)
= \(a^2+b^2-c^2+2ab\) (Ở đây ta gạch a2,b2,c2,2a)
_____________________________
\(a^2+b^2+c^2+2ac\) (Ở đây ta cũng gạch a2,b2,c2,2a)
=> Kết quả cuối của biểu thức là: \(\frac{b}{c}\)
Tíck cho mình nha