Rút gọn phân thức:
\(a,\dfrac{\left(a+b\right)^2-c^2}{a+b+c}\)
\(b,\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
Rút gọn:
\(\dfrac{\left(a+b\right)^3-c^3}{a+b+c}\)
\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
Cái đầu ko rút gọn được
Cái sau:
\(=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a-b+c}\)
Rút gọn:
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
với: c2+2ab-2ac-2bc=0; b\(\ne\)c; a+b\(\ne\)c
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)
\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)
Cho a, b, c khác nhau đôi một và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\). Rút gọn các biểu thức:
a) M= \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Rút gọn phân thức:
\(a,\dfrac{\left(a+b\right)^2-c^2}{a+b+c}\)
\(b,\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
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) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{(a^2+b^2+2ab)-c^2}{(a^2+c^2+2ac)-b^2}\)
\(=\frac{(a+b)^2-c^2}{(a+c)^2-b^2}=\frac{(a+b-c)(a+b+c)}{(a+c-b)(a+c+b)}=\frac{a+b-c}{a+c-b}\)
Rút gọn các biểu thức sau
b, \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2a\left(b-c\right)\)
\(=a^2-2a\left(b-c\right)+2a\left(b-c\right)\)
\(=a^2\)
Rút gọn các phân thức :
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}\) b.\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
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 ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
Rút gọn các phân thức sau :
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}\) b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
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) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Rút gọn các phân thức sau :
a) \(\dfrac{\left(a+b\right)^2-c}{a+b+c}\)
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
c)\(\dfrac{2x^2-7x^2-12x+45}{3x^3-19x^3+33x-9}\)
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
a: Sửa đề: \(\dfrac{\left(a+b\right)^2-c^2}{a+b+c}\)
\(=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
c: Sửa đề: \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\dfrac{2x^3+5x^2-12x^2-30x+18x+45}{3x^3-x^2-18x^2+6x+27x-9}\)
\(=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)}=\dfrac{2x+5}{3x-1}\)