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\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)
\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)
\(c.\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)
\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{3}\times\left[\left(\frac{-3}{4}+\frac{1}{4}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)
\(=\frac{7}{3}\times\frac{1}{2}\)
\(=\frac{7}{6}\)
\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)
\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)
\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)
\(=\frac{7}{8}\times\frac{44}{7}\)
\(=\frac{11}{2}\)
Câu 22: Hãy sắp xếp số sau theo thứ tự tăng dần: -3; +4; 7; -7; 0; -1.
A. – 7; - 3 ; - 1; 0 ; + 4 ; 7. B. 7; +4; 0; -1; -3; -7.
C.7; -7; +4; -3; -1; 0. D. 0; -1; -3; +4; - 7; 7.
Câu 23: Liệt kê các phần tử của mỗi tập hợp: P = { x ε Z ǀ -2 ≤ x < 4}
A. P = { -2; -1; 0; 1; 2; 3; 4}. B. P = {-1; 0; 1; 2; 3; 4}.
C. P = {-1; 0; 1; 2; 3}. D. P = {-2; -1; 0; 1; 2; 3}.
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
a) 5 : 3/4 - 4 4/5 : 3/4
= 5 . 4/3 - 24/5 . 4/3
= (5 - 24/5) . 4/3
= 1/5 × 4/3
= 4/15
b) -3/5 . 2/7 + (-3/7) . 3/5 + (-3/7)
= (-3/7) . (2/5 + 3/5 + 1)
= (-3/7) . 2
= -6/7
c) [(-4 2/7) . 7/11 + 7/11 . (5 1/3)] . 5 - 5 2/3
= (-30/7 . 7/11 + 7/11 . 16/3) . 5 - 17/3
= (-30/11 + 112/33) . 5 - 17/3
= 2/3 . 5 - 17/3
= 10/3 - 17/3
= -7/3
d) 5/39 . [(7 4/5) . (1 2/3) + (8 1/3) . (7 4/5)]
= 5/39 . (39/5 . 5/3 + 25/3 . 39/5)
= 5/39 . 39/5 . (5/3 + 25/3)
= 1 . 10
= 10
=
Bài 2:
a: =>3/4x=-3/5-1/2=-11/10
\(\Leftrightarrow x=\dfrac{-11}{10}:\dfrac{3}{4}=\dfrac{-11}{10}\cdot\dfrac{4}{3}=-\dfrac{44}{30}=-\dfrac{22}{15}\)
b: \(\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{19}{12}\)
=>7/4x=19/12
=>x=19/21
c: \(\Leftrightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
=>-4/3x=-1/3-1/6=-1/2
=>x=1/2:4/3=1/2x3/4=3/8
Bài 1
\(\dfrac{1}{7}:\dfrac{5}{17}-\dfrac{3}{2}.\left(\dfrac{1}{6}-\dfrac{7}{12}\right)\)
\(\dfrac{1}{7}.\dfrac{17}{5}-\dfrac{3}{2}.\left(-\dfrac{5}{12}\right)\)
\(\dfrac{17}{35}-\left(-\dfrac{5}{8}\right)\)
\(\dfrac{17}{35}+\dfrac{5}{8}\)
\(\dfrac{311}{280}\)
\(-\dfrac{3}{4}+\dfrac{3}{7}-1\\ =-\dfrac{9}{28}-1\\ =-\dfrac{37}{28}.\)