\(16^5:8^3\)

\(=\left(2^4\right)^5:\left(2^3\right)^3\)

\(=2^{4\cdot5}:2^{3\cdot3}\)

\(=2^{20}:2^9\)

\(=2^{20-9}\)

\(=2^{11}\)

\(=2048\)

a, A = { n \(\in N\)| 1 =< n =< 5 } 

b, B = { n \(\in\)N| 0 =< n =< 4 } 

c, C = { n \(\in N\)| 1 =< n =< 4 } 

d, D = { \(n\in N\)| 0 =< n =< 3, 2n+2 } 

e, E = { \(n\in\)N| 0 =< n =< 24, 2n+1 }

f, F = { n \(\in\)N| 1 =< n =< 9, 11n }

Bài 2:

a) \(A=\left\{x\in N|1\le x\le5\right\}\) 

b) \(B=\left\{x\in N|0\le x\le4\right\}\)

c) \(C=\left\{x\in N|1\le x\le4\right\}\)

d) \(D=\left\{x=2k|k\in N;0\le k\le4\right\}\)

e) \(E=\left\{x=2k+1|k\in N;0\le k\le21\right\}\)

f) \(F=\left\{x=11k|k\in N;1\le k\le9\right\}\)

Bài 4:

\(B=\left[\left(-\dfrac{1}{2}\right):0,07-\dfrac{1}{5}:0,07\right]\left(\dfrac{3}{2}\cdot2,5+1,5\cdot\dfrac{15}{2}\right)\)

\(=\left[\left(-\dfrac{1}{2}\right)\cdot\dfrac{100}{7}-\dfrac{1}{5}\cdot\dfrac{100}{7}\right]\left(\dfrac{3}{2}\cdot\dfrac{5}{2}+\dfrac{3}{2}\cdot\dfrac{15}{2}\right)\)

\(=\dfrac{100}{7}\cdot\left(-\dfrac{1}{2}-\dfrac{1}{5}\right)\cdot\dfrac{3}{2}\left(\dfrac{5}{2}+\dfrac{15}{2}\right)\)

\(=\dfrac{100}{7}\cdot-\dfrac{7}{10}\cdot\dfrac{3}{2}\cdot\dfrac{20}{2}\)

\(=-10\cdot15\)

\(=-150\)

Chọn C 

Bài 5:

\(A=\left(\dfrac{20}{21}-1\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)

\(=\left(1-\dfrac{1}{21}-1\right):\left(\dfrac{41\cdot101}{42\cdot101}-1\right):\left(\dfrac{63\cdot10101}{64\cdot10101}-1\right)\)

\(=\left(-\dfrac{1}{21}\right):\left(\dfrac{41}{42}-1\right):\left(\dfrac{63}{64}-1\right)\)

\(=\left(-\dfrac{1}{21}\right):\left(1-\dfrac{1}{42}-1\right):\left(1-\dfrac{1}{64}-1\right)\)

\(=\left(-\dfrac{1}{21}\right):\left(-\dfrac{1}{42}\right):\left(-\dfrac{1}{64}\right)\)

\(=\left(-\dfrac{1}{21}\right)\cdot\left(-42\right)\cdot\left(-64\right)\)

\(=\left(-\dfrac{1}{21}\cdot-42\right)\cdot64\)

\(=2\cdot64\)

\(=128\)

Chọn A 

\(27^3:3^4\)

\(=\left(3^3\right)^4:3^4\)

\(=3^{3\cdot4}:3^4\)

\(=3^{12}:3^4\)

\(=3^{12-4}\)

\(=3^8\)

\(\dfrac{27^3}{3^4}\)

\(=\dfrac{\left(3^3\right)^3}{3^4}\)

\(=\dfrac{3^{3\cdot3}}{3^4}\)

\(=\dfrac{3^9}{3^4}\)

\(=3^{9-4}\)

\(=3^5\)

`#3107.101107`

`b,`

\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\) 

Ta thấy: `8 = 7 + 1 = x + 1`

Thay `8 = x + 1` vào, ta có:

\(x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)

\(=x-5=7-5=2.\)

Ta có:

+) \(\dfrac{103}{105}=\dfrac{105}{105}-\dfrac{2}{105}=1-\dfrac{2}{105}=1-\dfrac{6}{315}\)

+) \(\dfrac{205}{208}=\dfrac{208}{208}-\dfrac{3}{208}=1-\dfrac{3}{208}=1-\dfrac{6}{416}\)

Vì \(\dfrac{6}{315}>\dfrac{6}{416}\) nên \(1-\dfrac{6}{315}< 1-\dfrac{6}{416}\) hay \(\dfrac{103}{105}< \dfrac{205}{208}\)

Ta có:

\(\left\{{}\begin{matrix}1-\dfrac{103}{105}=\dfrac{2}{105}\\1-\dfrac{105}{208}=\dfrac{2}{208}\end{matrix}\right.\)

Vì \(105< 208\Rightarrow\dfrac{2}{105}>\dfrac{2}{208}\Rightarrow\dfrac{103}{105}< \dfrac{203}{208}\)

\(\left|2x-3\right|-x=\left|2-x\right|\) 

TH1: \(\dfrac{3}{2}\le x\le2\)

\(\Rightarrow\left(2x-3\right)-x=2-x\)

\(\Leftrightarrow x-3=2-x\)

\(\Leftrightarrow2x=5\)

\(\Leftrightarrow x=\dfrac{5}{2}\left(ktm\right)\) 

TH2: \(x>2\) 

\(\Rightarrow\left(2x-3\right)-x=x-2\)

\(\Leftrightarrow x-3=x-2\)

\(\Leftrightarrow0=1\) (vô lý)  

TH3: \(x< \dfrac{3}{2}\) 

\(\Rightarrow\left(3-2x\right)-x=2-x\)

\(\Leftrightarrow3-3x=2-x\)

\(\Leftrightarrow3-2=-x+3x\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

Vậy: .... 

2 bài nào đấy bạn?

Bạn cần hỏi điều gì?

Ta có: \(2\) là số tự nhiên \(\Rightarrow2^{32}\) là số tự nhiên 

\(\Rightarrow2^{32}+1\) là số tự nhiên 

Tổng số bao gạo là:

   11 + 15 + 25 = 51 (bao gạo)

Kho thứ hai chứa tổng cộng số kg gạo là:

   15 x 45 = 675 (kg gạo)

Tổng số kg gạo có ở cả ba kho là:

   550 + 675 + 1325 = 2550 (kg gạo)

TB mỗi bao gạo chứa số kg gạo là:

  2550 : 51 = 50 (kg gạo)

Vì đây vẫn là toán lớp 5 nên mình vẫn nên có đáp số để bài rõ ràng hơn em nhé.