x + 16 \(⋮\) x + 1
x + 11 \(⋮\) x + 1
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27 tháng 1 2022
\(A=\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{55}{334}=\dfrac{11}{334}\)
\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{20}{21}\)
PA
4
21 tháng 10 2023
em ghi rõ cộng trừ nhân chia đi em
chị thấy đề lạ quá '* ' cái này lf cái j vậy em
HT
3
20 tháng 7 2015
5B = 1/11 - 1/16 + 1/16 - 1/21 + 1/21 - 1/26 +.........+ 1/56 - 1/61 +1/61 - 1/66
5B = 1/11 - 1/66
5B = 5/66
B = 1/66
NN
3
VV
2
15 tháng 8 2016
\(C=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{2011.2016}\)
\(5C=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{2011.2016}\)
\(5C=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{2011}-\frac{1}{2016}\)
\(5C=1-\frac{1}{2016}\)
\(5C=\frac{2015}{2016}\)
\(C=\frac{2015}{2016}:5\)
\(C=\frac{403}{2016}\)
15 tháng 8 2016
Đặt A = \(\frac{1}{1\times6}+\frac{1}{6\times11}+\frac{1}{11\times16}+...+\frac{1}{2011\times2016}\)
\(A\times5=\frac{5}{1\times6}+\frac{5}{6\times11}+\frac{5}{11\times16}+...+\frac{5}{2011\times2016}\)
\(A\times5=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{2011}-\frac{1}{2016}\)
\(A\times5=\frac{1}{1}-\frac{1}{2016}\)
\(A=\frac{2015}{2016}\times\frac{1}{5}\)
\(A=\frac{2015}{10080}=\frac{403}{2016}\)
31 tháng 8 2023
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
31 tháng 8 2023
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
5 tháng 1 2019
a, 1 x 2 x 3 x ... x 8 x 9 - 1 x 2 x 3 x ... x 8 - 1 x 2 x 3 x ... x 7 x 8 x 8
= 1 x 2 x 3 x ... x 8 ( 9 - 1 - 8 ) = 1 x2 x 3 x ... x 8 . 0 = 0
KV
30 tháng 9 2023
A = 1 × 6 + 6 × 9 + 11 × 16 + 16 × 19 + 21 × 26
= 6 + 54 + 176 + 304 + 546
= 1086
Vậy chữ số tận cùng của A là 6
1 tháng 10 2023
A=(...6)+(...4)+(.....6)+(....4)+(....6)
A=(....6)
vậy chữ số tận cùng của A bằng 6
vì x + 16 ⋮ x + 1
x+1⋮ x+1
=>(x+16)-(x+1)\(⋮x+1\)
=>\(15⋮x+1\)
=>(x+1)\(\inƯ\left(15\right)\) ={\(\pm1;\pm3;\pm5;\pm15\) }
ta có bảng
vậy x\(\in\left\{-16;-6;-4;-2;0;2;4;14\right\}\)
x + 11 ⋮ x + 1
vì \(x+1⋮x+1\)
=>\(\left(x+11\right)-\left(x+1\right)⋮\left(x+1\right)\)
=>\(\left(x+11-0x-1\right)⋮\left(x+1\right)\)
=> \(10⋮x+1\)
=>\(x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
ta có bảng sau:
9
vậy\(x\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)