| x + 2017 | - x 2017 = x
- | x + 2006 | - 2006 = x
x+1/2017+x+2/2016+x+3/2015=x+10/2008+x+11/2007+x+12/2006
A,2 x 32 x 12 + 4 x 6 x 41 + 8 x 27 x 3
B,(2006 x 2005^2016 - 2005^2016) : 2005^2017
`#3107.101107`
A,
\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)
B,
\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)
2 x 3 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 3 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 3 + 24 x 42 + 24 x 27
= 24 x ( 3 + 42 + 27 )
= 24 x 72
= 1728
Nho tick cho minh nhe ban😁
tính xyz
(x-1)^2006 + (2y-1)^2016+/x+2y-z/^2017=0
Ta có :
\(\left(x-1\right)^{2006}\ge0\)
\(\left(2y-1\right)^{2016}\ge0\)
\(\left(x+2y-z\right)^{2017}\ge0\)
Mà \(\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}\)\(+|x+2y-z|^{2017}\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2006}=0\\\left(2x-1\right)^{2016}=0\\|x+2y-z|^{2017}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\2y-1=0\\x+2y-z=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=1\\2y=1\\1-1-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy ...
Ta có :
\(\left(x-1\right)^{2006}\ge0\)
\(\left(2y-1\right)^{2016}\ge0\)
\(\left|x+2y-z\right|^{2017}\ge0\)
Mà \(\left(x-1\right)^{2006}+\left(2x-1\right)^{2016}+\left|x+2y-z\right|^{2017}=0\)
Suy ra : \(\hept{\begin{cases}\left(x-1\right)^{2006}=0\\\left(2x-1\right)^{2016}=0\\\left|x+2y-z\right|^{2017}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\2y-1=0\\x+2y-z=0\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\2y=1\\1+1-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy \(x=1\)\(;\)\(y=\frac{1}{2}\) và \(z=2\)
Chúc bạn học tốt ~
giải phương trình\(\frac{x}{2001}+\frac{x-3}{2009}+\frac{x+5}{2017}=\frac{x}{2012}+\frac{x-6}{2006}+\frac{x+10}{2022}\)
Khó quá, bạn nào giúp mk đi
1. Tìm 2 chữ số tận cùng
A= 2007 x 2009 x .....x 2017- 2002 x 2004 x 2006 x 2008
Thanks
tìm giá trị nhỏ nhất của A :
A = 2017-2006 ; (35-x)
Giải phương trình :
a) \(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
b) \(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
cho x = \(\frac{1}{2}.\left(7^{2016^{2017}-}23^{2006^{2001}}\right)\) Chứng minh x chia hết cho 3
Tính bằng cách thuận tiện nhất
a, 2017 * 4 + 2017 * 62 + 2017 * 35 - 2017
b. 2006 * 42 + 2006 * 39 + 2006 * 21 - 2006 * 2
Ai nhanh và đúng mik k và tick cho!!!!
2017*4+2017*62+2017*35-2017
=2017*(4+62+35-1)
=2017*100
=201700
2006*42+2006*39+2006*21-2006*2
=2006*(42+39+21-2)
=2006*100
=200600
k mik đi
a) 2017*4+2017*62+2017*35-2017
=2017*(4+62+35)-2017*1
=2017*(101-1)
=2017*100
=201700
b)2006*42+2006*39+2006*21-2006*2
=2006*(42+39+21-2)
=2006*100
=200600