PV
2
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
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)
CC
20 tháng 2 2020
\(ĐKXĐ:x\ne\pm2\)
\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left[\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right]\)
\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{\left(x-2\right)\left(x+2\right)}=\frac{2x}{\left(x+2\right)^2}.\frac{-\left(x-2\right)\left(x+2\right)}{x}\)
\(=\frac{-2\left(x-2\right)}{x+2}\)
BA
20 tháng 2 2020
\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2x}{\left(x+2\right)^2}\cdot\frac{\left(x-2\right)\left(x+2\right)}{x+4}\)
\(\Leftrightarrow\frac{2x^2-4x}{\left(x+2\right)\left(x+4\right)}\)
M
2
H9
HT.Phong (9A5)
CTVHS
2 tháng 8 2023
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1+x-18+x+2}{x-5}\)
\(=\dfrac{3x-15}{x-5}\)
\(=\dfrac{3\left(x-5\right)}{x-5}\)
\(=\dfrac{3}{1}\)
\(=3\)
2 tháng 8 2023
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
22 tháng 4 2020
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
NN
22 tháng 4 2020
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
10 tháng 9 2017
Ta có:\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{2}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\frac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{16+16}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
\(\frac{4}{x-1}+\frac{2}{1-x}+\frac{x}{x-1}\)
\(=\frac{4}{x-1}-\frac{2}{x-1}+\frac{x}{x-1}\)
\(=\frac{4-2+x}{x-1}\)
\(=\frac{2+x}{x-1}\)
P/s tham khảo nha
Thank bn nha