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(x – 5)2016 = (x – 5)2018
=> (x – 5)2018 – (x – 5)2016 = 0
=> (x – 5)2016.[(x – 5)2 – 1] = 0
=> x – 5 = 0 hoặc x – 5 = 1 hoặc x – 5 = -1
=> x = 5 hoặc x = 6 hoặc x = 4 (Thỏa mãn x ∈ N).
Vậy x ∈ {4; 5; 6}.
\(\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\)
<=> \(\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\)
<=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)<=>\(\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)
Vậy x\(\in\){4,5,6}
<=>
\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)
\(\Rightarrow\left(x-5\right)^{2016}\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{2016}=0\\\left(x-5\right)^2=1-0=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
\(\frac{x+4}{2018}+\frac{x+5}{2017}+\frac{x+6}{2016}+\frac{x+7}{2015}=-4\)
\(\Rightarrow\left(\frac{x+4}{2018}+1+\frac{x+5}{2017}+1+\frac{x+6}{2016}+1+\frac{x+7}{2015}+1\right)=-4+4=0\)
\(\Rightarrow\frac{x+2022}{2018}+\frac{x+2022}{2017}+\frac{x+2022}{2016}+\frac{x+2022}{2015}=0\)
\(\Rightarrow\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(\Rightarrow x+2022=0\Leftrightarrow x=-2022\)
1 a/ B = 2016 . 2016 - 2014 . 2018
=\(2016^2-\left(2016-2\right).\left(2016+2\right)\)
= \(2016^2-2016.\left(-2+2\right)\\\)
=\(2016^2-2016.0 \)
= \(2016^2\)
1: Tính
a) B = 2016 x 2016 - 2014 x 2018
= (2016 x 2016) - (2014 x 2018)
= 4064256 - 4064252
= 4
b) 327 x 412 + 400 / 328 x 412 -12
= (327 x 412 + 400) / ( 328 x 412 - 12)
= 135124 / 135124
= 1
c) ( 1234 x 567 - 667 ) : ( 567 + 1234 x 5)
= (1234 x 567 - 667) : (567 + 1234 x 566)
= [1234 x 566 + (1234 - 667)] / (1234 x 566 + 567)
= (1234 x 566 + 567) / (1234 x 566 + 567)
= 1
để thiếu bạn ơi
Nghiệm?
\(\left(x-5\right)^{2016}+\left(x-5\right)^{2018}=0\\ \Rightarrow\left(x-5\right)^{2016}\left[1+\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\1+\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x\in\varnothing\end{matrix}\right.\)
Yêu cầu bài là gì bạn nhỉ?