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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 ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+c+b\right)}\)
\(=\frac{a+b-c}{a+c-b}\)
Bạn sai đề nên mik sửa và làm luôn nha
Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\Rightarrow ab+ac+bc=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ac-ab\end{cases}}\)
Nên \(\frac{a^2}{a^2+2bc}=\frac{a^2+ab+bc+ac}{a^2+bc-ac-ab}=\frac{\left(a+c\right)\left(a+b\right)}{\left(a-c\right)\left(a-b\right)}\)
\(\frac{b^2}{b^2+2ac}=\frac{b^2+ab+bc+ac}{b^2+ac-ab-bc}=\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}\)
\(\frac{c^2}{b^2+2ab}=\frac{c^2+ab+ac+bc}{b^2+ab-ac-bc}=\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{\left(a+b\right)\left(a+c\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a+b\right)\left[\left(a+c\right)\left(b-c\right)+\left(b+c\right)\left(c-a\right)\right]+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a+b\right)\left(2bc-2ac\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{-2c\left(a+b\right)\left(a-b\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left[-2c\left(a+b\right)+\left(b+c\right)\left(c+a\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(-a^2+ab+c^2-bc\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Vậy \(P=1\)
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
\(\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\frac{a+b-c}{a-b+c}\)
\(\text{nhận xét: ta có hằng đẳng thức:}\)
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
đó đến đây bạn làm tiếp
\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2a\left(b-c\right)\)
\(=a^2-2a\left(b-c\right)+2a\left(b-c\right)\)
\(=a^2\)
cho \(c^2+2ab-2ac-2bc\)
rút gọn biểu thức \(P=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
Ta có :
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a+b\right)^2-2ab-c^2+2ab}{\left(a+c\right)^2-2ac-b^2+2ac}.\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)
\(=\frac{a+b-c}{a+c-b}\)
\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)
\(=\frac{a+b-c}{a+c-b}\left(a+b+c\ne0\right)\)
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.
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 ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
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) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)