\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)

\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)

\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)

\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)

=\(7^5.5.\left(-27\right)=-2268945\)

A = 7⁵*5*-27

a: \(=\left(\dfrac{5^4}{5^2\cdot7^2}\right)^{15}:\left(\dfrac{5^6}{2\cdot7\cdot31}\right)^7\)

\(=\dfrac{5^{30}}{7^{30}}:\dfrac{5^{42}}{2^7\cdot7^7\cdot31^7}\)

\(=\dfrac{5^{30}}{7^{30}}\cdot\dfrac{2^7\cdot7^7\cdot31^7}{5^{42}}=\dfrac{2^7\cdot31^7}{7^{23}\cdot5^{12}}\)

b: \(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\)

\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-7\cdot4\right)}{5^{29}\cdot2^8\cdot7^{48}}=5\cdot\left(-27\right)=-135\)

\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{3^2\cdot2^2\cdot5^2\cdot7^{48}}\)

\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1+7\cdot4\right)}{3^2\cdot2^2\cdot5^2\cdot7^{48}}=\dfrac{5^{28}\cdot2^6\cdot12}{3^2}=\dfrac{5^{28}\cdot2^8}{3}\)

\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)

\(A=\frac{7^{49}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(A=\frac{7^{48}.5^{30}.2^8\left(1-28\right)}{5^{29}.2^8.7^{48}}\)

\(A=5.\left(-27\right)\)

\(A=-135\)

Tìm x:

\(\dfrac{4-x}{6-x}\)=\(\dfrac{x-3}{x-8}\)\(\Rightarrow\)(4-x)(x-8)=(6-x)(x-3)

\(\Rightarrow\)12x-x²-32=9x-x²-18

\(\Rightarrow\)3x=14\(\Rightarrow\)x=\(\dfrac{14}{3}\).

\(\dfrac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

=\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

=\(\dfrac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)

=5.(1-7.2²) = 5.(1-28) = 5.(-27) = -135

\(\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

\(=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)

\(=5.\left(1-7.4\right)\)

\(=5.\left(1-28\right)\)

\(=5.\left(-27\right)=-135\)

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)

\(\dfrac{\left(-14\right).15}{21.\left(-10\right)}\\ =\dfrac{-7.2.3.5}{7.3.-2.5}\\=\dfrac{7.2.3.5}{7.2.3.5}\\ =1\)

b)

\(\dfrac{5.7-7.9}{7.2+6.7}\\ =\dfrac{7\left(5-9\right)}{7\left(2+6\right)}\\ =\dfrac{-4}{8}\\ =\dfrac{-2.2}{2.4}\\ =-\dfrac{1}{2}\)

c)

\(\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}\\ =\dfrac{-7.3+2.-7.2}{7\left(-5-2\right)}\\ =\dfrac{-7\left(3+4\right)}{7.-7}\\ =\dfrac{7}{7}\\ =1\)

a) S = 1.2 + 2.3 + 3.4 + ... + 99.100

S có thể được viết lại thành:

S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)

= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98

= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)

Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:

S = n(n+1)(2n+1)/6

Với n = 99, ta có:

S = 99.100.199/6 = 331650

Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:

S = n(n+1)/2

Với n = 98, ta có:

S = 98.99/2 = 4851

Do đó, S = 331650 - 4851 = 326799

b) B = 4924.12517.28−530.749.45529.162.748

B có thể được viết lại thành:

B = (4924.12517.28) / (530.749.45529.162.748)

B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)

B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529

B = 108 / 45529

c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101

C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101

C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)

C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)

C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)

d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018

D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^