Tính
\(\dfrac{3}{x^2+6x+9_{ }}+\dfrac{2}{6x-x-9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
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
DT
1
12 tháng 2 2022
Sửa đề: \(\dfrac{3}{x^2+6x+9}-\dfrac{3}{x^2-6x+9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
\(=\dfrac{3x^2-18x+27-3x^2-18x-27+x^2+30x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{x^2-6x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}=\dfrac{\left(x-9\right)\left(x+3\right)}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{\left(x-9\right)}{\left(x^2-9\right)\left(x-3\right)}\)
29 tháng 11 2022
a: \(=\dfrac{x^3-x^2+x-1}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x+1\right)}-\dfrac{3x}{\left(x-2\right)\left(x+1\right)}+\dfrac{2x+5}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+1\right)\left(x+1\right)-x^2+4x-4-3x^2-6x+2x+5}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{x^4-1-4x^2+1}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}=\dfrac{x^2\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}\)
=x^2/x+1
b: Sửa đề: \(\dfrac{19x^2-30x+9}{2x^3+54}-\dfrac{x-3}{2x^2+6x}-\dfrac{3x^2}{2x^2-6x+18}\) \(=\dfrac{19x^2-30x+9}{2\left(x+3\right)\left(x^2-3x+9\right)}-\dfrac{x-3}{2x\left(x+3\right)}-\dfrac{3x^2}{2\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-\left(x-3\right)\left(x^2-3x+9\right)-3x^3\left(x+3\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-3x^4-9x^3-\left(x^3-3x^2+9x-3x^2+9x-27\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-30x^2+9x-x^3+6x^2-18x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-24x^2-9x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
13 tháng 12 2018
\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)
\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{1}{2\left(x-3\right)}\)
\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)
\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)
\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)
\(=\frac{12}{x}\)
7 tháng 12 2022
a: \(=\dfrac{x^2+3x+9}{2\left(x+5\right)}\cdot\dfrac{\left(x+5\right)}{\left(x-3\right)\left(x^2+3x+9\right)}=\dfrac{1}{2\left(x-3\right)}\)
b: \(=\dfrac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\dfrac{6x^2+37x+6+6x^2-37x+6}{x}\cdot\dfrac{1}{x^2+1}=\dfrac{12}{x}\)
19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
\(x=\dfrac{3+\sqrt{5}}{2}\Rightarrow2x-3=\sqrt{5}\Rightarrow4x^2-12x+9=5\)
\(\Rightarrow4x^2-12x+4=0\Rightarrow x^2-3x+1=0\)
\(\Rightarrow P=\left[10\left(x^2-3x+1\right)+1\right]^2+\dfrac{\left[2\left(x^2-3x+1\right)+1\right]^{10}}{x^3\left(x^2-3x+1\right)-1}=1^2+\dfrac{1^2}{0-1}=...\)
Akai Haruma Nguyễn Huy Tú Nguyễn Huy ThắngHồng Phúc NguyễnPhạm Hoàng Giang......và nhiều bạn nữa giúp mik vs
\(\dfrac{3}{x^2+6x+9}+\dfrac{2}{6x-x^2-9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
\(=\dfrac{3}{\left(x+3\right)^2}+\dfrac{-2}{\left(x-3\right)^2}+\dfrac{x^2+30x-27}{x^4-9x^2-9x^2+81}\)
\(=\dfrac{3}{\left(x+3\right)^2}-\dfrac{2}{\left(x-3\right)^2}+\dfrac{x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{3\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}-\dfrac{2\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\dfrac{x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{3x^2-18x+27-2x^2-12x-18+x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2x^2-18}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2\left(x^2-9\right)}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x^2-9}\)