KV
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KV
0
HN
0
13 tháng 12 2018
\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)
\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{1}{2\left(x-3\right)}\)
\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)
\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)
\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)
\(=\frac{12}{x}\)
3 tháng 3 2022
a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)
\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)
b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x^2-3x}\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 8 2023
a) \(\dfrac{2x}{x^2-6x+9}+\dfrac{x-2}{x-3}\) (ĐK: \(x\ne3\))
\(=\dfrac{2x}{\left(x-3\right)^2}+\dfrac{x-2}{x-3}\)
\(=\dfrac{2x}{\left(x-3\right)^2}+\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{2x+x^2-2x-3x+6}{\left(x-3\right)^2}\)
\(=\dfrac{x^2-3x+6}{x^2-6x+9}\)
b) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
12 tháng 3 2020
\(\frac{6}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
12 tháng 3 2020
\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\left(x\ne\pm3\right)\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x^2+15x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
P2
20 tháng 12 2019
Bài 1:
a)
\(9x^2-49=0\)
\(9x^2-49+49=0+49.\)
\(9x^2=49\)
\(\frac{9x^2}{9}=\frac{49}{9}\)
\(x^2=\frac{49}{9}\)
\(x=\sqrt{\frac{49}{9}}\)
\(x=\frac{\sqrt{49}}{\sqrt{9}}\)
\(x=\frac{7}{3}\)hay \(x=2,33333...\)
b)
\(\left(x-1\right)\left(x+2\right)-x-2=0.\)
\(x^2+x-2-x-2.\)
\(x^2+\left(x-x\right)-\left(2+2\right)=\)\(0\)
\(x^2-4=0\)
\(x=\sqrt{4}\)
\(x=2\)
Bài 2:
a)
\(\frac{x}{x}-3+9-\frac{6x}{x^2}-3x.\)
\(=1-3+9-\frac{6x}{x^2}-3x.\)
\(=1-3+9-\frac{6}{x}-3x.\)
\(=7-\frac{6}{x}-3x\)
b)
\(6x-\frac{3}{x}\div4x^2-\frac{1}{3x^2}\)
\(=6x-\frac{3}{x}\div\frac{4}{1}x^2-\frac{1}{3x^2}.\)
\(=6x-\frac{3}{x}\times\frac{1}{4}x^2-\frac{1}{3x^2}\)
\(=6x-\frac{3x^2}{x4}-\frac{1}{3x^2}\)
\(=6x-\frac{3x}{4}-\frac{1}{3x^2}\)
\(=\frac{6x}{1}-\frac{3x}{4}-\frac{1}{3x^2}\)
\(=\frac{72x^3-36x^3-12x^2}{12x^2}\)
\(=\frac{36-12x^2}{12x^2}\)
Cứu với mn ơi huhu !!