Toán

a: \(A=4+2^2+2^3+...+2^{20}\)

=>\(2A=8+2^3+2^4+...+2^{21}\)

=>\(2A-A=2^{21}+2^{20}+...+2^4+2^3+8-2^{20}-2^{19}-...-2^3-2^2-4\)

\(=2^{21}+8-2^2-4=2^{21}\)

=>\(A=2^{21}\) là lũy thừa của 2

b:

\(B=3+3^2+3^3+...+3^{100}\)

=>\(3B=3^2+3^3+...+3^{101}\)

=>\(2B=3^{101}-3\)

=>\(2B+3=3^{101}\) là lũy thừa của 3

1: \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{7n^2-4}{n-5}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(7-\dfrac{4}{n^2}\right)}{n\left(1-\dfrac{5}{n}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(7-\dfrac{4}{n^2}\right)}{1-\dfrac{5}{n}}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7}{1}=7>0\end{matrix}\right.\)

2: 

\(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-3+\dfrac{2}{n^2}\right)}{n\left(1-\dfrac{2}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(-3+\dfrac{2}{n^2}\right)}{1-\dfrac{2}{n}}\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{-3+\dfrac{2}{n^2}}{1-\dfrac{2}{n}}=-\dfrac{3}{1}=-3< 0\end{matrix}\right.\)

x-5 chia hết cho 7

=>\(x-5\in\left\{0;7;14;...\right\}\)

=>\(x\in\left\{5;12;19;26;33;40;47;54;...\right\}\)

mà 0<=x<50

nên \(x\in\left\{5;12;19;26;33;40;47\right\}\)

60 chia hết cho x

=>\(x\inƯ\left(60\right)\)

=>\(x\in\left\{1;-1;2;-2;3;-3;4;-4;5;-5;6;-6;10;-10;12;-12;15;-15;20;-20;30;-30;60;-60\right\}\)

mà x>=6

nên \(x\in\left\{6;10;12;15;20;30;60\right\}\)

60 chia hết cho x 

\(\Rightarrow x\inƯ\left(60\right)\)

Mà: \(Ư\left(60\right)=\left\{1;2;3;4;5;6;10;12;15;30;60\right\}\)

Lại có: \(x\ge6\)

\(\Rightarrow x\in\left\{6;10;12;15;30;60\right\}\)

x - 5 chia hết cho 7 nên:

\(\Rightarrow x-5\in B\left(7\right)=\left\{0;7;14;21;28;35;42;49;56;...\right\}\)

\(\Rightarrow x\in\left\{5;12;19;26;33;40;47;...\right\}\)

Mà: \(0\le x\le50\)

\(\Rightarrow x\in\left\{5;12;19;26;33;40;47\right\}\)

\(x-5⋮7\)

=>\(x-5\in\left\{0;7;14;21;28;35;42;49;56;...\right\}\)

=>\(x\in\left\{5;12;19;26;33;40;47;54;...\right\}\)

mà \(0< x< =50\)

nên \(x\in\left\{5;12;19;26;33;40;47;54\right\}\)

36 chia hết cho x - 2 

\(\Rightarrow x-2\inƯ\left(36\right)\)

Mà: \(Ư\left(36\right)=\left\{1;2;3;4;6;9;12;18;36\right\}\)

\(\Rightarrow x\in\left\{3;4;5;6;8;11;14;20;38\right\}\)

a: Xét tứ giác ABOC có

\(\widehat{OBA}+\widehat{OCA}=180^0\)

=>OBAC là tứ giác nội tiếp

b: Xét (O) có

DB,DM là tiếp tuyến

=>DB=DM và OD là phân giác của \(\widehat{BOM}\left(1\right)\)

Xét (O) có

EM,EC là tiếp tuyến

=>EM=EC và OE là phân giác của \(\widehat{MOC}\left(2\right)\)

\(C_{ADE}=AD+DE+AE\)

\(=AB-BD+DM+ME+AC-CE\)

\(=AB+AC=2AB\)

c: \(\widehat{DOE}=\widehat{DOM}+\widehat{EOM}\)

\(=\dfrac{1}{2}\left(\widehat{BOM}+\widehat{COM}\right)=\dfrac{1}{2}\cdot\widehat{BOC}\)

`#3107.101107`

a)

`5/9 + 4/9 * 3/7 + 4/9 * 4/7`

`= 5/9 + 4/9 * (3/7 + 4/7)`

`= 5/9 + 4/9 * 7/7`

`= 5/9 + 4/9 * 1`

`= 5/9 + 4/9`

`= 1`

b)

`(1/3 + 3/8 - 7/12) \div 1/8`

`= (17/24 - 7/12) \div 1/8`

`= 1/8 \div 1/8`

`= 1`

c)

`(1 + 2/3 - 5/4) - (1 - 5/4) + (2022 - 2/3)`

`= 1 + 2/3 - 5/4 - 1 + 5/4 + 2022 - 2/3`

`= (1 + 2022) + (2/3 - 2/3) + (-5/4 + 5/4)`

`= 2023`

Vì \(x⋮6\) nên \(\Rightarrow x\in B\left(6\right)=\left\{0;6;12;18;24;30;36;42;48;...\right\}\)

Mà:\(20\le x\le42\)

\(\Rightarrow x\in\left\{24;30;36;42\right\}\)