Rút gọn phân thức
\(\dfrac{3^{3x}-3^{3y}}{3^x+3^y}\)
\(\dfrac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
UN
2
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 9 2023
\(b,\dfrac{x^3+3x^2-2}{x^3+3x+4}=\dfrac{x^3+x^2+2x^2+2x-2x-2}{x^3+x^2-x^2-x+4x+4}\\ =\dfrac{x^2\left(x+1\right)+2x\left(x+1\right)-2\left(x+1\right)}{x^2\left(x+1\right)-x\left(x+1\right)+4\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^2+2x-2\right)}{\left(x+1\right)\left(x^2-x+4\right)}=\dfrac{x^2+2x-2}{x^2-x+4}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 9 2023
\(a,\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\dfrac{\left(x^2-y^2\right)+\left(3x-3y\right)}{x^2-y^2}\\ =\dfrac{\left(x-y\right)\left(x+y\right)+3\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y+3\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+3}{x+y}\)
18 tháng 4 2023
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
MH
15 tháng 9 2023
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
L
18 tháng 2 2021
\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)
\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)
18 tháng 2 2021
a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)
\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)
\(=\dfrac{7y}{8x}\)
26 tháng 10 2023
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
12 tháng 4 2021
Đặt bthuc = A nhé
ĐKXĐ : \(2x\ne3y\)
\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)
\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)
\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)
Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
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) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)
22 tháng 7 2023
`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`.
12 tháng 10 2021
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
b) Ta có: \(\dfrac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)
\(=\dfrac{4^{2m}-4^{2n}}{4^n+4^m}\)
\(=\dfrac{\left(4^m+4^n\right)\left(4^m-4^n\right)}{4^n+4^m}\)
\(=4^m-4^n\)