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2) -3(4 - 7) + 5(-3 + 2)
= -3.4 + 3.7 - 5.3 + 5.2
= -12 + 21 -15 + 10
= 31 - 27
= 4
4) -5(2 - 7) + 4(2 - 5)
= -5.2 + 5.7 + 4.2 - 4.5
= -10 + 35 + 8 - 20
= 38 - 30
= 8
3: \(=20-12-8+12=20-8=12\)
5: \(=-18-42-21-35=-116\)
3: \(=-15+18-12+8=-27+26=-1\)
2: \(=-12+21-15+10=9-5=4\)
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
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, \(E=\frac{2^{9011}-2}{3}\)
Ta có:
Đáp án A là: \(\dfrac{3}{4}+\dfrac{7}{2}=\dfrac{3}{4}+\dfrac{14}{4}=\dfrac{17}{4}\)
Đáp án B là: \(\dfrac{3}{4}-\dfrac{7}{2}=\dfrac{3}{4}-\dfrac{14}{4}=-\dfrac{11}{4}\)
Đáp án C là: \(-\dfrac{3}{4}+\dfrac{7}{2}=-\dfrac{3}{4}+\dfrac{14}{4}=\dfrac{11}{4}\)
Đáp án D là: \(-\dfrac{3}{4}-\dfrac{7}{2}=-\dfrac{3}{4}-\dfrac{14}{4}=-\dfrac{17}{4}\)
Vậy đáp án B là chính xác
7)=-5.(-1)-7.2
=5-14
=-9
8)=7.(-2) - 9.(-5)
= -14-(-45)
=31
9)= -8.(-1)+7.4
=8+28
=36
10)= -2.(-2)+4.2
=4+8
=12
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
=
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)