Lời giải:
$a-11b+3c\vdots 17$

$\Rightarrow 2(a-11b+3c)\vdots 17$

$\Rightarrow 2a-22b+6c\vdots 17$

$\Rightarrow 2a-5b+6c-17b\vdots 17$

$\Rightarrow 2a-5b+6c\vdots 17$ (đpcm)

`#3107.101107`

a)

\(49\cdot52+48\cdot132-48\cdot83\\ =49\cdot52+48\cdot\left(132-83\right)\\ =49\cdot52+48\cdot49\\ =49\cdot\left(52+48\right)\\ =49\cdot100\\ =4900\)

b)

\(42\cdot53+47\cdot186-114\cdot47\\ =42\cdot53+47\cdot\left(186-114\right)\\ =42\cdot53+47\cdot72\\ =2226+3384\\ =5610\)

thằng nào có tiền, thì nạp vào roblox donate cho ta. ít ,thì 5 robux. nhiều , thì 1k robux

Ta có:

5⁶⁰ⁿ = (5³)²⁰ⁿ = 125²⁰ⁿ

2¹⁴⁰ⁿ = (2⁷)²⁰ⁿ = 128²⁰ⁿ

3¹⁰⁰ⁿ = (3⁵)²⁰ⁿ = 243²⁰ⁿ

Do 125 < 128 < 243

125²⁰ⁿ < 128²⁰ⁿ < 243²⁰ⁿ

Vậy 5⁶⁰ⁿ < 2¹⁴⁰ⁿ < 3¹⁰⁰ⁿ

Không có đủ cơ sở để đưa ra kết luận này bạn nhé.

{50:5-45:5}x7

={10-9}x7

=1x7

=7

d,

=6^2x10:{780:[1000-15,625+35x14]}

=6^2x10:{780:[1000-15,625+490]}

=6^2x10:{780:1474,375}

=6^2x10:0,52903772785

=36x10:0,52903772785

=360:0,52903772785

=608,480769232

khum bt câu d, làm đúng chưa?

chắc sai á

a)2¹¹:{1026-[3⁴+1]:41}

=2¹¹:{1026-82:41}

=2¹¹:2¹¹

=2^11-11=2

b) (3⁵+13):4⁴*(2022*2021-4082419)

=256:4⁴*4043

=0*4043

=0

chú thích:

*=dấu nhân

3²ⁿ = (3²)ⁿ = 9ⁿ

2³ⁿ = (2³)ⁿ = 8ⁿ

Do 9 > 8 nên 9ⁿ > 8ⁿ

Vậy 3²ⁿ > 2³ⁿ

------------

5³⁶ = (5³)¹² = 125¹²

11²⁴ = (11²)¹² = 121¹²

Do 125 > 121 nên 125¹² > 121¹²

Vậy 5³⁶ > 11²⁴

`#3107.101107`

a)

\(3^{2n}\) và \(2^{3n}\)

Ta có:

\(3^{2n}=3^{2\cdot n}=\left(3^2\right)^n=9^n\\ 2^{3n}=2^{3\cdot n}=\left(2^3\right)^n=8^n\)

Vì \(9>8\Rightarrow9^n>8^n\Rightarrow3^{2n}>2^{3n}\)

Vậy, \(3^{2n}>2^{3n}\)

b)

\(5^{36}\) và \(11^{24}\)

Ta có:

\(5^{36}=5^{12\cdot3}=\left(5^3\right)^{12}=125^{12}\\ 11^{24}=11^{12\cdot2}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125>121\Rightarrow125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

Vậy, \(5^{36}>11^{24}.\)

\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\\ \Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

`#3107.101107`

\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\)

\(\Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\cdot\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy, \(x\in\left\{4;5;6\right\}.\)

`#3107.101107`

\(\dfrac{7^{40}\cdot5-7^{39}\cdot8}{7^{39}\cdot3^3}\)

\(=\dfrac{7^{39}\cdot\left(7\cdot5-8\right)}{7^{39}\cdot3^3}\\ =\dfrac{7\cdot5-8}{3^3}\\ =\dfrac{35-8}{27}\\ =\dfrac{27}{27}\\ =1\)