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\(a,\dfrac{7}{35},\dfrac{18}{54},\dfrac{-15}{125},\dfrac{-4}{25}\)
Các thừa số đã tối giản : \(\dfrac{-4}{25}\)
\(\dfrac{7}{35}=\dfrac{7:7}{35:7}=\dfrac{1}{5}\) , \(\dfrac{18}{54}=\dfrac{18:18}{54:18}=\dfrac{1}{3}\)
\(\dfrac{-15}{125}=\dfrac{-15:5}{125:5}=\dfrac{-3}{25}\)
\(b,\dfrac{27}{45},\dfrac{21}{28},\dfrac{8}{14},\dfrac{18}{-60},\dfrac{-270}{360}\)
Các thừa số đã tối giản là : ko có
\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\) , \(\dfrac{21}{28}=\dfrac{21:7}{28:7}=\dfrac{3}{4}\)
\(\dfrac{8}{14}\)\(=\dfrac{8:2}{14:2}=\dfrac{4}{7}\) , \(\dfrac{18}{-60}=\dfrac{18:6}{-60:6}=\dfrac{3}{-10}=\dfrac{-3}{10}\)
\(\dfrac{-270}{360}=\dfrac{-270:90}{360:90}=\dfrac{-3}{4}\)
\(c,\dfrac{3.4+3.7}{6.5+9}\) = \(\dfrac{3.\left(4+7\right)}{30+9}\) = \(\dfrac{3.11}{39}\) = \(\dfrac{3.11}{3.13}=\dfrac{11}{13}\)
\(\dfrac{-63}{81},\dfrac{9.6}{9.35},\dfrac{7.2+8}{2.14.5}\)
Các p/s đã tối giản : ko có
\(\dfrac{-63}{81}=\dfrac{-63:9}{81:9}=\dfrac{-7}{9}\) , \(\dfrac{9.6}{9.35}=\dfrac{6}{35}\)
\(\dfrac{7.2+8}{2.14.5}=\dfrac{14+8}{28.5}=\dfrac{22}{140}=\dfrac{11}{70}\)
Bài 2.
A = -3/5 + ( -2/5 + 2 )
A = -3/5 + ( -2/5 + 10/5 )
A = -3/5 + 8/5
A = 5/5
A = 1
--------------------------------------------------------
B = 3/7 + ( -1/5 + -3/7 )
B = 3/7 + ( -7/35 + -15/35 )
B = 3/7 + ( -22/35 )
B = 15/35 + ( -22/35 )
B = -1/5
-----------------------------------------------------
C = ( -5/24 + 0,75 + 7/12 ) : ( -2 . 1/8 )
C = ( -5/24 + 3/4 + 7/12 ) : ( -1/4 )
C = 9/8 : ( -1/4 )
C = 9/8 . ( -4 )
C = -9/2
Bài 3 .
a) 4/7 - x = 1/2 . x + 2/7
<=> -x - x = 1/2 - 4/7 + 2/7
<=> -2x = 3/14
<=> x = 3/14 . ( -1/2 )
<=> x = -3/28
Vậy x = -3/28
b) x : 3 1/5 = 1 1/2
<=> x : 16/5 = 3/2
<=> x = 3/2 . 16/5
<=> x = 24/5
Vậy x = 24/5
c) x . 3/4 = -1 5/8
<=> x . 3/4 = -13/8
<=> x = -13/8 . 4/3
<=> x = -13/6
Vậy x = -13/6
d)
\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)
e)
\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)
f)
\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)
\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)
\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)
\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)
\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)
\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)
\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)
\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)
Đặt A = 1 + 3 + 5 + ... + 97 + 99
Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)
Tổng A bằng: (99 + 1) . 50 : 2 = 2500
Thay A = 2500 vào biểu thức (1), ta được:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\cdot...\left(1+\dfrac{1}{2499}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot...\cdot\dfrac{2500}{2499}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{50\cdot50}{49\cdot51}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot50}{1\cdot2\cdot3\cdot...\cdot49}\cdot\dfrac{2\cdot3\cdot...\cdot50}{3\cdot4\cdot...\cdot51}\)
\(=\dfrac{50}{1}\cdot\dfrac{2}{51}=\dfrac{100}{51}\)
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
Lời giải:
$A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2023}}$
$2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2022}}$
$2A-A=2-\frac{1}{2^{2023}}$
$A=2-\frac{1}{2^{2023}}$
q=1/3; u1=2/3
\(S_{100}=\dfrac{\dfrac{2}{3}\cdot\left(\dfrac{1}{3^{100}}-1\right)}{\dfrac{1}{3}-1}=-\dfrac{1}{3^{100}}+1=\dfrac{-1+3^{100}}{3^{100}}\)
\(\dfrac{5.6}{9.35}=\dfrac{5.3.2}{3.3.5.7}=\dfrac{2}{3.7}=\dfrac{2}{21}\)
\(\dfrac{7.2+8}{2.14.5}=\dfrac{7.2+8.2}{2.7.2.5}=\dfrac{\left(7+8\right).2}{2.7.2.5}=\dfrac{15.2}{2.7.2.5}=\dfrac{3.5.2}{2.7.2.5}=\dfrac{3}{14}\)
1,\(\dfrac{5.6}{9.35}\) = \(\dfrac{5.2.3}{3.3.5.7}\) = \(\dfrac{2}{3.7}\)
2, \(\dfrac{7.2+8}{2.14.5}\) = \(\dfrac{7.2+2.4}{2.14.5}\) = \(\dfrac{2\left(7+4\right)}{2\left(14.5\right)}\) = \(\dfrac{2.11}{2.70}\)= \(\dfrac{11}{70}\)