Rút gọn phân thức :
a, \(\dfrac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
b, \(\dfrac{m^4-m}{2m^2+2m+2}\)
c, \(\dfrac{ab^2+a^3-a^2b}{a^3+b^4}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
\(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)
\(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)
\(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)
Đề lỗi rồi chứ mình ko rút gọn đc nữa
c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)
\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)
a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)
b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)
c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)
d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)
`a, (3x^2y)/(2xy^5)`
`= (3x)/(2y^4)`
`b, (3x^2-3x)/(x-1)`
`= (3x(x-1))/(x-1)`
`= 3x`
`c, (ab^2-a^2b)/(2a^2+a)`
`= (b(a-b))/((2a+1))`
`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
a) \(\dfrac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\dfrac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}=\dfrac{1}{\left(a+b\right)\left(c+d\right)}\)
b) \(\dfrac{m^4-m}{2m^2+2m+2}=\dfrac{m\left(m^3-1\right)}{2\left(m^2+m+1\right)}=\dfrac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\dfrac{m\left(m-1\right)}{2}\)
c) \(\dfrac{ab^2+a^3-a^2b}{a^3+b^3}=\dfrac{a\left(b^2+a^2-ab\right)}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\dfrac{a}{a+b}\)
bạn ơi giúp minh 1 câu này đc o :
\(\dfrac{xy+1-x-y}{y+z-1-yz}\)
Vs cả câu c ý mẫu là a\(^3\)b + b\(^4\) nha chứ o phải b\(^3\)