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a) Đặt \(A=\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b\right)^2}{a+b}-\frac{c^2}{c}=a+b-c\)
b)Đặt \(B=\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{a+b-c}{a+c-b}\)
Auto giải thích thêm câu b) (để tránh bị các thành phần spammer bắt bẻ)
\(\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{a+b-c}{a+c-b}\) vì:
\(\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left[\left(a+b\right)-c\right]\left[\left(a+b\right)+c\right]}{\left[\left(a+c\right)-b\right]\left[\left(a+c\right)+b\right]}=\frac{a+b-c}{a+c-b}\)
cho mình hỏi bạn biết làm chưa nếu rồi thì giúp mình được không ạ mình ko biết làm
Lời giải:
ĐK:............
Theo hằng đẳng thức đáng nhớ ta có:
a) \(\frac{(a+b)^2-c^2}{a+b+c}=\frac{(a+b-c)(a+b+c)}{a+b+c}=a+b-c\)
b) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{(a^2+b^2+2ab)-c^2}{(a^2+c^2+2ac)-b^2}\)
\(=\frac{(a+b)^2-c^2}{(a+c)^2-b^2}=\frac{(a+b-c)(a+b+c)}{(a+c-b)(a+c+b)}=\frac{a+b-c}{a+c-b}\)
a) \(\dfrac{ax+ay-bx-by}{ax-ay-bx+by}=\dfrac{a\left(x+y\right)-b\left(x+y\right)}{a\left(x-y\right)-b\left(x-y\right)}=\dfrac{\left(a-b\right)\left(x+y\right)}{\left(a-b\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a+c-b}\)
a) sai đề
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+c+b\right)}=\dfrac{a+b-c}{a-b+c}\)
c) xem lại đề có j ib lại tui
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+2ab+b^2+2bc+c^2+2ca=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
*\(a^2+2bc=a^2+bc-ca-ab=\left(a-c\right)\left(a-b\right)\)
Tương tự cho 2 cái còn lại.
Ta có:
\(C=\dfrac{a^2}{a^2+bc-ab-ca}+\dfrac{b^2}{b^2+ac-ab-bc}+\dfrac{c^2}{c^2+ab-bc-ca}\)
\(C=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}+\dfrac{b^2}{\left(a-b\right)\left(b-c\right)}+\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\)
Tới đây cứ việc quy đồng mẫu là được.
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow ab+bc+ca=0\)
\(C=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
\(=\dfrac{a^2}{a^2+bc-ac-ab}+\dfrac{b^2}{b^2+ac-ba-bc}+\dfrac{c^2}{c^2+ab-ca-cb}\)
\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\left(\dfrac{a^2}{\left(a-b\right)\left(c-a\right)}+\dfrac{b^2}{\left(a-b\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\left(\dfrac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)=1\)
\(C=\dfrac{a^2+b^2-c^2+2ab}{a+b+c}\)
\(C=\dfrac{\left(a^2+2ab+b^2\right)-c^2}{a+b+c}\)
\(C=\dfrac{\left(a+b\right)^2-c^2}{a+b+c}\)
\(C=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{a+b+c}\)
\(C=a+b-c\)
a,\(C=\dfrac{a^2+b^2-c^2+2ab}{a+b+c}=\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{a+b+c}=a+b-c\)b, \(D=\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a-b+c\right)\left(a+b+c\right)}=\dfrac{a+b-c}{a-b+c}\)