A) 150 - 100 : (15 . 6 - 40)

= 150 - 100 : (90 - 40)

= 150 - 100 : 50

= 150 - 2

= 148

B) 3.(63 - 58)² + 2⁷ : 2⁴ . 2⁰

= 3.5² + 2³

= 75 + 8

= 83

A) \(150-100:\left(15\cdot6-40\right)\)

\(=150-100:\left(90-40\right)\)

\(=150-100:50\)

\(=150-2\)

\(=148\)

B) \(3\left(63-58\right)^2+2^7:2^4\cdot2^0\)

\(=3\cdot5^2+2^7:2^4\)

\(=3\cdot25+2^3\)

\(=75+8\)

\(=83\)

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

5\(x\) + \(x\) = 30

6\(x\)        = 30

  \(x\)    = 30: 6

  \(x\)    = 5

Ta thấy : \(A\text{=}2.4.6.8.....20\)

Trong A là các tích được nhân bởi các số chẵn từ 2 đến 20 .

Mà trong A có hạng tử 8 \(\Rightarrow A⋮8\)

Trong A lại có hạng tử là 20 \(\Rightarrow A⋮20\)

Vậy...........

Có nha

A = \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)

A = \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

A = \(\dfrac{1}{3}\) - \(\dfrac{1}{9}\)

A = \(\dfrac{2}{9}\)

B = \(\dfrac{1}{5.6}\)+ \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+...+\(\dfrac{1}{23.24}\)+\(\dfrac{1}{24.25}\)

B = \(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{23}\)-\(\dfrac{1}{24}\)+ \(\dfrac{1}{24}\)-\(\dfrac{1}{25}\)

B = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)

B = \(\dfrac{4}{25}\)

\(21\cdot7^2-11\cdot7^2+90\cdot7^2+7^2\cdot125\cdot16\)

\(=7^2\left(21-11+90+2000\right)\)

\(=7^2\cdot2100\)

\(=102900\)

= 7 mũ 2 . ( 21 - 11 + 90 ) + 49.125.16

= 7 mũ 2 . 100 + 49.125.16

= 4900 + 49.125.16

= 4900 + 98000

= 102900

\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)  (sửa \(-9^9\rightarrow.9^9\))

\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{2^{29}.3^{18}\left(15-9\right)}{2^{28}.3^{18}\left(15-14\right)}\)

\(=2.6=12\)

6^31=186

3^23=69

2020^3010=6080200

   Mik ko chắc nx.Số to quá mik ko lm đầy đủ đc nên mik viết luôn kết quả nha,có j cho mik xl

ko sao cảm ơn bạn

\(A=1\cdot2\cdot...\cdot10\)

\(A=1\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot\left(2\cdot5\cdot10\right)\)

\(A=1\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot100⋮100\)

Vậy \(A⋮100\)

Lời giải:

$8\leq 2n\leq 128$

$\Rightarrow \frac{8}{2}\leq n\leq \frac{128}{2}$

$\Rightarrow 4\leq n\leq 64$

$\Rightarrow n\in\left\{4; 5; 6; ....; 63; 64\right\}$