NV
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
NM
1
BH
23 tháng 8 2019
Đặt \(\left(x^2+2x+2\right)^2=t\Rightarrow\frac{1}{t^2}+\frac{1}{\left(t+1\right)^2}=\frac{5}{4}\Leftrightarrow\frac{\left(t^2+2t+1\right)+t^2}{t^2\left(t+1\right)^2}=\frac{5}{4}\Leftrightarrow4\left(2t^2+2t+1\right)=5\left(t^4+2t^3+t^2\right)\) \(\Leftrightarrow8t^2+8t+4=5t^4+10t^3+5t^2\Leftrightarrow5t^4+10t^3-3t^2-8t-4=0\)\(\Leftrightarrow\left(t-1\right)\left(t+2\right)\left(5t^2+5t+2\right)=0\Leftrightarrow\hept{\begin{cases}t=1\\t=-2\end{cases}}\)
21 tháng 9 2019
a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)
\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)
\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)
\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow x^2-2.x.2+2^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=2\)
31 tháng 7 2019
\(A=\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10-5}\)
\(A=\frac{\left(2x+1\right)\left(2x+1\right)-\left(2x-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{10-5}\)
\(A=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{10-5}\)
\(A=\frac{\left(2x\right)^2+2.2x+1-\left(2x\right)^2+2.2x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4}{10-5}\)
\(A=\frac{\left(2x\right)^2+4x+1-\left(2x\right)^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{10-5}\)
\(A=\frac{\left[\left(2x\right)^2-\left(2x\right)^2\right]+\left(4x+4x\right)+\left(1-1\right)}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{10-5}\)
\(A=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{10-5}\)
\(A=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x}{5}\)
\(A=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}:\left(4x.5\right)\)
\(A=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}:20x\)
\(A=\frac{8x}{20x\left(2x-1\right)\left(2x+1\right)}\)
\(A=\frac{8}{20\left(2x-1\right)\left(2x+1\right)}\)
\(A=\frac{2}{5\left(2x-1\right)\left(2x+1\right)}\)
22 tháng 4 2020
Bài 1 :
Ta có :
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)
\(+\left(\frac{x+2013}{2011}+1\right)\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+4024=0\)
\(\Rightarrow x=-4024\)
22 tháng 4 2020
Bài 2 :
Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)
=> Phương trình trở thành
\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)
\(\Rightarrow5a^2+3a-8=0\)
\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)
Vì \(a\ge0\Rightarrow a=1\)
\(\Rightarrow x^2+2x+1=1\)
\(x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2,0\right\}\)
\(\frac{x+1}{2x-2}+\frac{2}{1-x^2}=\frac{x-1}{2x+2}\)
\(\frac{x+1}{2\left(x-1\right)}-\frac{2}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{2\left(x+1\right)}=0\)
ĐKXĐ: x \(\ne\) + 1
\(\frac{\left(x+1\right)^2-4-\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=0\)
\(\frac{x^2+2x+1-4-x^2+2x-1}{2\left(x-1\right)\left(x+1\right)}=0\)
\(\frac{4x-4}{2\left(x-1\right)\left(x+1\right)}=0\)
\(\frac{4\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=0\)
\(\frac{4}{2\left(x+1\right)}=0\)
\(\Rightarrow x\in\phi\)
ĐK : x khác 1 ; -1
Pt <=> (x + 1 )^2 - 4 = (x-1)^2
<=> x^2 +2x+ 1 - 4 = x^2 - 2x + 1
=> 4x = 4
=> x = 1 (loại )