Thực hiện các phép tính (tính nhanh nếu có thể) A = 3 4 . 8 9 . 15 16 ... 2499 2500
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=\(\frac{3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{49\cdot51}{50\cdot50}\)
=\(\frac{\left(2\cdot3\cdot...\cdot49\right)\cdot\left(3\cdot4\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...\cdot50\right)}\)
=\(\frac{51}{50\cdot2}=\frac{51}{100}\)
a: \(2\cdot25\cdot4\cdot50\)
\(=\left(2\cdot50\right)\cdot\left(25\cdot4\right)\)
\(=100\cdot100=10000\)
b: \(\left(-125\right)\cdot5\cdot\left(-16\right)\cdot\left(-8\right)\)
\(=-125\cdot5\cdot16\cdot8\)
\(=-\left(125\cdot8\right)\cdot\left(5\cdot16\right)\)
\(=-80\cdot1000=-80000\)
c: \(5^2\cdot3^3\cdot2\)
\(=25\cdot27\cdot2\)
\(=\left(25\cdot2\right)\cdot27=27\cdot50=1350\)
d: \(\left(-4\right)\cdot3^2\left(-5\right)^3\)
\(=\left(-4\right)\cdot\left(-125\right)\cdot9\)
\(=500\cdot9=4500\)
a) \(3.5^2-16:2^2\)
\(=3.25-16.4\)
\(=75-64\)
\(=11\)
b) \(17.85+15.17-120\)
\(=17.\left(85+15\right)-120\)
\(=17.100-120\)
\(=1700-120\)
\(=1580\)
c) \(\left(3^{15}.4+5.3^{15}\right):3^{16}\)
\(=\left[3^{15}.\left(4+5\right)\right]:3^{16}\)
\(=\left[3^{15}.9\right]:3^{16}\)
\(=\left[3^{15}.3^2\right]:3^{16}\)
\(=3^{17}:3^{16}\)
\(=3\)
a. 3.52 - 24 : 22 = 3.52- 22 = 3.25-4= 75-4 =71
b. 17.(85+15) - 120 = 17.100-120=1700-120=1580
c. |315.(4+5)| :316= (315.9) : 316= 315.32 : 316= 317:316=3
dấu | là ngoặc vuông bn nhé
a) Ta có \(A=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\dfrac{4\cdot6}{5\cdot5}\cdot...\cdot\dfrac{49\cdot51}{50\cdot50}\)
\(=\dfrac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\cdot\dfrac{4\cdot5\cdot6\cdot...\cdot51}{3\cdot4\cdot5\cdot...\cdot50}\)
= \(\dfrac{2}{50}\cdot17=\dfrac{17}{25}\)
b) Vì n nguyên nên 3n - 1 nguyên
Để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên thì 12 ⋮ ( 3n - 1 ) hay ( 3n - 1 ) ϵ Ư( 12 )
Ư( 12 ) = { \(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\) }
Lập bảng giá trị
3n - 1 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
n | \(\dfrac{2}{3}\) | 0 | 1 | \(\dfrac{-1}{3}\) | \(\dfrac{3}{4}\) | \(\dfrac{-2}{3}\) | \(\dfrac{5}{3}\) | -1 | \(\dfrac{7}{3}\) | \(\dfrac{-5}{3}\) | \(\dfrac{13}{3}\) | \(\dfrac{-11}{3}\) |
Vì n nguyên nên n ϵ { 0; 1; -1 }
Vậy n ϵ { 0; 1; -1 } để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên
Trả lời:
\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}\)
\(=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}\)
\(=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot50\cdot50}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot49\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...50\right)}\)
\(=\frac{1\cdot51}{50.2}=\frac{51}{100}\)