Câu 1: Cho a, b ∈ ℤ, b ≠ 0, x = ab.Nếu a, b khác dấu thì:
- A. x = 0;
- B. x> 0;
- C. x < 0;
- D. Cả B, C đều sai.
- Mn ơi giúp tôi ik!!! Cần gấp!!!
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1 người ăn hết số gạo đó trong:
\(3\cdot12=36\left(ngày\right)\)
9 người ăn hết số gạo đó trong:
\(36:9=4\left(ngày\right)\)
Đáp số: 4 ngày
A = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
3A = 1 - \(\dfrac{2}{3^{ }}\) + \(\dfrac{3}{3^2}\) - \(\dfrac{4}{3^3}\) + ... + \(\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)
3A+A = 1-\(\dfrac{2}{3^{ }}\)+\(\dfrac{3}{3^2}\)-\(\dfrac{4}{3^3}\)+...+\(\dfrac{99}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)+\(\dfrac{1}{3}-\dfrac{2}{3^2}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
4A = 1-(\(\dfrac{2}{3}\)-\(\dfrac{1}{3}\)) +(\(\dfrac{3}{3^2}\)-\(\dfrac{2}{3^2}\))-(\(\dfrac{4}{3^3}\)-\(\dfrac{3}{3^3}\))+...+(\(\dfrac{99}{3^{98}}\)-\(\dfrac{98}{3^{98}}\))-(\(\dfrac{100}{3^{99}}\)-\(\dfrac{99}{3^{99}}\))-\(\dfrac{100}{3^{100}}\)
4A = 1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{98}}\)-\(\dfrac{1}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)
12A = 3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+....+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)
12A+4A=3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)+1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+..+\(\dfrac{1}{3^{98}}\)-\(\dfrac{1}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)
16A = 3+(-1+1)+(\(\dfrac{1}{3}-\dfrac{1}{3}\))+...+(-\(\dfrac{1}{3^{98}}\)+\(\dfrac{1}{3^{98}}\))+(-\(\dfrac{100}{3^{99}}\)-\(\dfrac{1}{3^{99}}\)) - \(\dfrac{100}{3^{100}}\)
16A = 3 - \(\dfrac{101}{3^{99}}\) - \(\dfrac{100}{3^{100}}\)
16A = 3 - \(\dfrac{303}{3^{100}}\) - \(\dfrac{100}{3^{100}}\)
16A = 3 - \(\dfrac{403}{3^{100}}\)
A = \(\dfrac{3}{16}\) - \(\dfrac{403}{16.3^{100}}\) < \(\dfrac{3}{16}\) < \(\dfrac{3}{14}\) (đpcm)
\(a,\dfrac{7}{-9}+\dfrac{-1}{-9}=\dfrac{-7}{9}+\dfrac{1}{9}=\dfrac{-7+1}{9}=\dfrac{-6}{9}=\dfrac{-2}{3}\\ b,\dfrac{7}{-18}+\left(\dfrac{-5}{12}-\dfrac{13}{-18}\right)=\dfrac{-7}{18}-\dfrac{5}{12}+\dfrac{13}{18}=\left(\dfrac{13}{18}-\dfrac{7}{18}\right)-\dfrac{5}{12}\\ =\dfrac{6}{18}-\dfrac{5}{12}=\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{1.4-5}{12}=\dfrac{-1}{12}\\ c,5-\dfrac{-7}{8}+\dfrac{15}{-20}=5+\dfrac{7}{8}-\dfrac{3}{4}=\dfrac{5.8+7-3.2}{8}=\dfrac{40+7-6}{8}=\dfrac{41}{8}\)
a) \(\dfrac{7}{-9}+\dfrac{-1}{-9}=\dfrac{6}{-9}=\dfrac{-2}{3}\)
b) \(\dfrac{7}{-18}+\left(\dfrac{-5}{12}-\dfrac{13}{-18}\right)\)
\(=\dfrac{7}{-18}-\dfrac{5}{12}-\dfrac{13}{-18}\)
\(=\dfrac{-6}{-18}-\dfrac{5}{12}\)
\(=\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{4}{12}-\dfrac{5}{12}\)
\(=\dfrac{-1}{12}\)
c) \(5-\dfrac{-7}{8}+\dfrac{15}{20}\)
\(=5-\dfrac{-7}{8}+\dfrac{3}{4}\)
\(=5-\dfrac{-7}{8}+\dfrac{6}{8}\)
\(=5-\dfrac{-1}{8}=5+\dfrac{1}{8}\)
\(=\dfrac{41}{8}\)
1; (\(\dfrac{-4}{25}\)).(-\(\dfrac{-25}{8}\))
= \(\dfrac{-4.25}{25.4.2}\)
= \(-\dfrac{1}{2}\)
2; \(\dfrac{5}{-14}\).(\(\dfrac{-7}{10}\))
= \(\dfrac{5.\left(-7\right)}{2.5.\left(-7\right).2}\)
= \(\dfrac{1}{4}\)
3; \(\dfrac{-15}{4}\).(\(\dfrac{-16}{25}\))
= \(\dfrac{3.5.4.4}{4.5.5}\)
= \(\dfrac{12}{5}\)
4; 15. (- \(\dfrac{13}{10}\))
= 5.3.\(\dfrac{\left(-13\right)}{2.5}\)
= - \(\dfrac{39}{2}\)
Số lần cắt: 4,8:0,8 - 1 = 5 (lần)
Số lần nghỉ: 5 - 1 = 4 (lần)
Người đó cắt xong khúc gỗ trong: 5 x 4 + 4 x 2 = 28 (phút)
Đ.số:....
\(D=\left\{6;7;8;9;10;11\right\}\)
\(D=\left\{x|x\in N,5< x< 12\right\}\)
____
\(5\notin D\)
\(7\in D\)
\(17\notin D\)
\(0\notin D\)
\(10\in D\)
Bài 1:
a; A = \(\dfrac{n-7}{n-4}\) (n \(\in\) Z)
A là phân số ⇔ n \(\in\) Z; n - 4 ≠ 0 ⇒ n ≠ 4
Vậy A là phân số khi 4 ≠ n \(\in\) Z
b; A \(\in\) Z ⇔ n - 7 ⋮ n - 4
n - 4 - 3 ⋮ n - 4
3 ⋮ n - 4
n - 4 \(\in\) Ư(3) = {-3; -1; 1; 3}
Lập bảng ta có:
n-4 | -3 | -1 | 1 | 3 |
n | 1 | 3 | 5 | 7 |
Theo bảng trên ta có n \(\in\) {1; 3; 5; 7}
Vậy n \(\in\) {1; 3; 5; 7}
Bài 2:
1; \(\dfrac{2009.2010-1}{2009.2010}\)
= \(\dfrac{2009.2010}{2009.2010}-\dfrac{1}{2009.2010}\)
= 1 - \(\dfrac{1}{2009.2010}\)
\(\dfrac{2010.2011-1}{2010.2011}\)
= \(\dfrac{2010.2011}{2010.2011}-\dfrac{1}{2010.2011}\)
= 1 - \(\dfrac{1}{2010.2011}\)
Vì \(\dfrac{1}{2009.2010}>\dfrac{1}{2010.2011}\)
Vậy \(\dfrac{2009.2010-1}{2009.2010}< \dfrac{2010.2011-1}{2010.2011}\)