6-(x-\(\dfrac{1}{2}\))\(^2\)=(-2)\(^{2018}\):2\(^{2014}\)
Những câu hỏi liên quan
giải các phương trình:
\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\dfrac{x+2}{2020}+\dfrac{x+4}{2018}=\dfrac{x+6}{2016}+\dfrac{x+8}{2014}\)
Giúp tớ với, thầy réo tớ kinh lắm rồi!
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
vậy
Đúng 1
Bình luận (0)
16 tháng 7 2021 lúc 13:27
Bài 6: So sánha,dfrac{1}{2}+dfrac{1}{_{ }2^2}+dfrac{1}{2_{ }^3}+...+dfrac{1}{2^{2014}}và 1 b,dfrac{10^{2018}+5}{10^{2018}-8}và dfrac{10^{2019}+5}{10^{2019}-8}c,dfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{23.24.25}vàdfrac{1}{4}
Đọc tiếp
Bài 6: So sánh
a,\(\dfrac{1}{2}\)+\(\dfrac{1}{_{ }2^2}\)+\(\dfrac{1}{2_{ }^3}\)+...+\(\dfrac{1}{2^{2014}}\)và 1 b,\(\dfrac{10^{2018}+5}{10^{2018}-8}\)và \(\dfrac{10^{2019}+5}{10^{2019}-8}\)
c,\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{23.24.25}\)và\(\dfrac{1}{4}\)
Tìm x,biết:
x+2015/5 + x+2014/6 = x+2017/3 + x+2018/2
Hướng dẫn: x+2015/5+1 + x+2014/6+1 = x+2017/3+1 + x+2018/2+1
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
Đúng 0
Bình luận (0)
Tìm x: a) ( x - 2018) ^ 2014 = 1
b) (\(\dfrac{1}{3}\) x ) : \(\dfrac{2}{3}\) = \(^7_4:\dfrac{2}{5}\)
Trần Trung NguyênThiên HànUyen VuuyenAki TsukiHuỳnh Ngọc Diễm LyDƯƠNG PHAN KHÁNH DƯƠNGRồng Đom ĐómNgân HảiNguyenNguyễn Việt LâmMashiro ShiinaDƯƠNG PHAN KHÁNH DƯƠNGNguyễn Thanh HằngCẩm MịchNguyễn Huy TúAce LegonaMashiro ShiinaMysterious PersonPhương AnTrần Việt Linh
Đúng 0
Bình luận (0)
a) \(\left(x-2018\right)^{2014}=1\)
\(\Rightarrow\left(x-2018\right)^{2014}=1^{2014}\)
\(\Rightarrow\left[{}\begin{matrix}x-2018=1\\x-2018=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1+2018\\x=-1+2018\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=2017\end{matrix}\right.\)
b) \(\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\)
\(\Rightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}.\dfrac{2}{5}\)
\(\Rightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{10}\)
\(\Rightarrow\dfrac{1}{3}x=\dfrac{7}{10}.\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{3}x=\dfrac{7}{15}\)
\(\Rightarrow x=\dfrac{7}{15}:\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{7}{15}.3\)
\(\Rightarrow x=\dfrac{7}{5}\)
Đúng 0
Bình luận (0)
a) \(\left(x-2018\right)^{2014}=1\Leftrightarrow\)\(\left[{}\begin{matrix}x-2018=\sqrt[2014]{1}\\x-2018=-\sqrt[2014]{1}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2018=1\\x-2018=-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2019\\x=2017\end{matrix}\right.\)
Vậy S={2017;2019}
b) \(\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}.\dfrac{5}{2}\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{35}{8}\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{8}.\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{12}\Leftrightarrow x=\dfrac{35}{12}:\dfrac{1}{3}\Leftrightarrow x=\dfrac{35}{4}\)
Vậy S={\(\dfrac{35}{4}\)}
Đúng 0
Bình luận (0)
Help me :
Tìm x :\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
Ta có:
\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)
Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)
Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
Đúng 0
Bình luận (0)
\frac{x-10}{2010}+\frac{x-8}{2012}+\frac{x-6}{2014}+\frac{x-4}{2016}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2016}{4}+\frac{x-2014}{6}+\frac{x-2012}{8}+\frac{x-2010}{10}
Tìm x biết :
a) \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
b) \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
c) (2x - 1 )2 = ( 2x - 1 )2018
d) ( x - 1 )x + 2 = ( x - 1 )x + 4
e) ( 2x - 3 )2 = 144
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
Đúng 0
Bình luận (0)
\(\dfrac{x+2014}{2}+\dfrac{2x+4028}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)
\(\dfrac{x+2014}{2}+\dfrac{2\left(x+2014\right)}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)
\(\left(x+2014\right)\left(\dfrac{1}{2}+\dfrac{2}{7}\right)=\left(x+2014\right)\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(\left(x+2014\right)\dfrac{11}{14}=\left(x+2014\right)\dfrac{11}{30}\)
Dấu ''=''↔x=-2014
Đúng 3
Bình luận (0)
Với x\(\ne-1\) \(\left(\dfrac{x^2+2x+2}{x+1}\right)^{2018}=a_0+a_1x+a_2x^2+...+a_kx^{2018}+\dfrac{b_1}{x+1}+\dfrac{b_2}{\left(x+1\right)^2}+...+\dfrac{b_{2018}}{\left(x+1\right)^{2018}}.\). Tính: S=\(\sum\limits^{2018}_{k=1}bx\)