\(\dfrac{x+2}{2014}\)+\(\dfrac{x+1}{2015}\)=\(\dfrac{x+2001}{15}\)+\(\dfrac{2014}{12}\)
Giúp mình với, mình cảm ơn nhiều ạ
giải phương trình
a) \(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
b) \(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
giải chi tiết giúp e ạ;-;
a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100
các bạn giúp mình câu này với
giải phương trình sau: \(\dfrac{x+2}{x-2}\)-\(\dfrac{2}{x^2-2x}\)=\(\dfrac{1}{x}\)
mình cảm ơn nhiều ạ.
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x^2-2x\)
\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)
Cho mình sửa lại nhé:
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Cho A = \(\dfrac{2015}{2014^2+1}+\dfrac{2015}{2014^2+2}+\dfrac{2015}{2014^3+3}+....+\dfrac{2015}{2014^2+2014}\)
Chứng minh rằng A không là số nguyên dương
Các bạn ơi , giúp mình với T T
Tìm x biết
\(\dfrac{2x-1}{-12}\)=\(\dfrac{48}{1-2x}\)
Giúp mình vs ạ, mình cảm ơn nhiều
=>(2x-1)^2=24^2
=>2x-1=24 hoặc 2x-1=-24
=>x=-23/2 hoặc x=25/2
\(\dfrac{2x-1}{-12}=\dfrac{48}{1-2x}\) (ĐK: \(x\ne\dfrac{1}{2}\))
\(\Leftrightarrow\dfrac{2x-1}{12}=\dfrac{48}{2x-1}\)
\(\Leftrightarrow-12\cdot48=\left(2x-1\right)\left(2x-1\right)\)
\(\Leftrightarrow567=\left(2x-1\right)^2\)
\(\Leftrightarrow24^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=-24\\2x-1=24\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-23\\2x=25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(tm\right)\\x=\dfrac{25}{2}\left(tm\right)\end{matrix}\right.\)
\(\dfrac{x-2014}{4}+\dfrac{x-2015}{3}=\dfrac{x-13}{2005}+\dfrac{x-14}{2004}\)
<=>\(\left(\dfrac{x-2014}{4}-1\right)+\left(\dfrac{x-2015}{3}-1\right)=\left(\dfrac{x-13}{2005}-1\right)+\left(\dfrac{x-14}{2004}-1\right)\)
<=>\(\dfrac{x-2018}{4}+\dfrac{x-2018}{3}=\dfrac{x-2018}{2005}+\dfrac{x-2018}{2004}\)
<=>\(\left(x-2018\right).\left[\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2005}-\dfrac{1}{2004}\right]=0\)
<=> \(x-2018=0\)
=>x=2018
Vậy S= {2018}
Chúc bạn học tốt!
#Yuii
giúp mình với ạ
Bài 1: giải các PT:
a, \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
b, \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
c, \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
d, \(\dfrac{x+\dfrac{2\left(3-x\right)}{5}}{14}-\dfrac{5x-4\left(x-1\right)}{24}=\dfrac{7x+2+\dfrac{9-3x}{5}}{12}+\dfrac{2}{3}\)
\(e,\dfrac{x-\dfrac{3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}}{ }\)
a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
haizzz bệnh lười lại lên cơn r
Cho biểu thức A=\(=\dfrac{2014}{1-x}+\dfrac{2014}{1+x}+\dfrac{4028}{1+x^2}+\dfrac{8056}{1+x^4}+\dfrac{16112}{1+x^8}+2,1314\)
Giúp mình với .Mh cần gấp
Có :
A = (2014/1-x + 2014/1+x) + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 4028/1-x^2 + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 8056/1-x^4 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 16112/1-x^8 + 16112/1+x^8 + 2,1314
= 32224/1-x^16 + 2,1314
Tk mk nha
Đề bài là gì vậy bạn
Sửa lại đề đi rùi báo cho mk để mk làm cho
Nhớ đó nha
Aaaaaaaa.......Sorry mình thiếu đề bài phần thiếu là Rút gọn rrooif tính giá trị của A khi x=1,1.Mh không cố í
Tìm x
a)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
b)\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
CẢM ƠN CÁC BẠN NHÌU ^_^
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}-\dfrac{x+2018}{2016}-\dfrac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow xx+2018=0\Leftrightarrow x=-2018\)
Vậy x = -2018
Giúp mình câu này với ạ
\(\dfrac{1}{x}\)+\(\dfrac{1}{x+2}\)+\(\dfrac{x-2}{x^2+2x}\)
Cảm ơn!
A nên đăng vào mấy h mà nhìu ng onl í
chứ h này ko còn ai đou ạ
\(=\dfrac{x+2+x+x-2}{x\left(x+2\right)}=\dfrac{3x}{x\left(x+2\right)}=\dfrac{3}{x+2}\)