Giải phương trình: x^2/(x^2+2x+2) + x^2/(x^-2x+2)-4(x^2-5)/(x^4+4)= 322/65
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
KT
14 tháng 2 2018
\(ĐKXĐ:\) \(\forall x\in Z\)
\(\frac{x^2}{x^2+2x+2}+\frac{x^2}{x^2-2x+2}-\frac{4\left(x^2-5\right)}{x^4+4}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{x^2\left(x^2-2x+2\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}+\frac{x^2\left(x^2+2x+2\right)}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}-\frac{4\left(x^2-5\right)}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{2x^4+10}{x^4+4}=\frac{322}{65}\)
\(\Rightarrow\)\(65\left(2x^4+10\right)=322\left(x^4+4\right)\)
\(\Leftrightarrow\)\(130x^4+650=322x^4+1288\)
\(\Leftrightarrow\)\(192x^4=-638\) (vô lý)
Vậy pt vô nghiệm
P/S:mk lm bừa thôi, đúng thì you tham khảo, sai thì báo mk biết nha
19 tháng 6 2022
\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow130x^4+1300=322x^4+1288\)
\(\Leftrightarrow-192x^4=-12\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
=>x=1/2 hoặc x=-1/2
TN
15 tháng 6 2019
a/ \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
<=> \(\frac{\left(x+1\right)^2}{\left(x+1\right)^2+1}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2+2}=\frac{7}{6}\left(1\right)\)
đặt \(\left(x+1\right)^2=a\left(a>0\right)\)
=> \(\left(1\right)\)<=> \(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
<=> \(\frac{a\left(a+2\right)+\left(a+1\right)^2}{\left(a+1\right)\left(a+2\right)}=\frac{7}{6}\)
<=> \(\frac{2a^2+4a+1}{a^2+3a+2}=\frac{7}{6}\)
<=> \(6\left(2a^2+4a+1\right)=7\left(a^2+3a+2\right)\)
<=> \(5a^2+3a-8=0\)
<=> \(5a^2-5a+8a-8=0\)
<=> \(\left(5a+8\right)\left(a-1\right)=0\)
<=> \(a=\frac{-8}{5}\left(h\right)a=1\)
mà \(a>0\)
=> \(a=1\)
=> \(\left(x+1\right)^2=1\)
=> \(x+1=1\left(h\right)x+1=-1\)
=> \(x=0\left(h\right)x=-2\)
vậy ......
chúc bn học tốt
15 tháng 6 2019
Xét x = 0 và x = -2 , thay vào ta được \(VT=VP\)
Xét x > 0 :
\(VT=\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=1-\frac{1}{x^2+2x+2}+1-\frac{1}{x^2+2x+3}\)
\(=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)>\frac{7}{6}=VP\) ( loại )
Xét x < -2 :
\(VT=2-\left(\frac{1}{x\left(x+2\right)+2}+\frac{1}{x\left(x+2\right)+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{7}{6}=VP\) ( loại )
Xét -2 < x < 0 :
\(VT=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{-2}+1\right)=\frac{3}{2}>\frac{7}{6}=VP\) ( loại )
Vậy ...
12 tháng 5 2018
bt2.
A=[2(4x^2+4x+5)-2]/(4x^2+4x+5)
=2-2/[(4x+1)^2+4]
A>=2-2/4=3/2
khi x=-1/4
18 tháng 7 2021
\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)
\(< =>2x-7=0< =>x=\dfrac{7}{2}\)
b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)
\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)
\(< =>-25x-11=0\)
\(< =>x=-0,44\)
15 tháng 5 2023
a: =>4x^2-24x+36-4x^2+4x-1<10
=>-20x<10-35=-25
=>x>=5/4
b: =>x(x^2-25)-x^3-8<=3
=>x^3-25x-x^3-8<=3
=>-25x<=11
=>x>=-11/25
ĐKXĐ : \(\forall x\)
Ta có : \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65\left(2x^4+20\right)=322\left(x^4+4\right)\)
\(\Leftrightarrow130x^4+1300=322x^4+1288\)
\(\Leftrightarrow192x^4-12=0\)
\(\Leftrightarrow x^4=\dfrac{12}{192}\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
\(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{1}{x^2+2+2x}+\dfrac{1}{x^2+2-2x}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{2x^2+4}{x^4+4}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+4x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65x^4+650=161x^4+644\)
\(\Leftrightarrow96x^4=6\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Rightarrow x=\pm\dfrac{1}{2}\)