Các bạn giúp mk với tối nay nộp rùi 

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a, \(A=3a.2.b-a.432b-4ab\)

\(=6ab-432ab-4ab=-430ab\)

b, \(A=-430ab=\left(-430\right).\frac{1}{229}.\frac{1}{433}=\frac{-430}{229.433}\)

~ So sad :( !! ~

\(A=\frac{31}{60}\)

I thinks so ! Sad

Đặt \(A=\frac{\left(5^2.6^{11}.16^2+6^2.12^6.15^2\right).10}{2.6^{12}.10^4-81^2.960^3}\)

\(A=\frac{\left(5^2.2^{11}.3^{11}.2^8+2^2.3^2.3^3.2^{12}.3^2.5^2\right).10}{2.2^{12}.3^{12}.2^4.5^4-9^4.2^{18}.3^3.5^3}\)

\(A=\frac{5^2.2^{11}.3^{11}.2^8.2.5+2^2.3^2.2^{12}.3^25^2.2.5}{2^{15}.\left(2^2.3^{12}.5^4-9^4.2^3.3^3.5^3\right)}\)

\(A=\frac{2^{15}\left(5^3.2^5.3^{11}+3^4.5^3\right)}{2^{15}.\left(2^2.3^{12}.5^4-9^4.2^3.3^3.5^3\right)}\)

\(A=\frac{5^3.2^5.3^{11}+3^4.5^3}{2^2.3^{12}.5^4-9^4.2^3.3^3.5^3}\)

\(A=\frac{5^3\left(2^5.3^{11}+3^4\right)}{5^3\left(2^2.3^{12}.5-9^4.2^3.3^3\right)}\)\(A=\frac{2^5.3^{11}+3^4}{2^2.3^{12}.5-9^4.2^3.3^3}\)

\(A=\frac{3^4.\left(2^5.3^8.5+1\right)}{3^4\left(2^2.3^8.5-3^4.2^3.3^3\right)}\)

\(A=\frac{2^5.3^8.5+1}{2^2.3^8.5-3^4.2^3.3^3}\)

Cậu phân tích từ từ

a) \(\frac{2.7.13}{26.35}=\frac{2.7.13}{2.13.7.5}=\frac{1}{5}\)

b) \(\frac{3.5.11.13}{33.35.37}=\frac{3.5.11.13}{3.11.7.5.37}=\frac{13}{259}\)

c) \(\frac{23.5-23}{4-27}=\frac{23.\left(5-1\right)}{-23}=\frac{4}{-1}=-4\)

d) \(\frac{1717-101}{2828+404}=\frac{101.17-101}{404.7+404}=\frac{101.\left(17-1\right)}{404.\left(7+1\right)}=\frac{101.16}{404.8}=\frac{101.2.8}{101.4.8}=\frac{1}{2}\)

\(\frac{23.5-23}{4-27}\)=\(\frac{23.\left(5-1\right)}{-23}\)=\(\frac{92}{-23}\)

Câu 1 Tính 

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)

Câu 2 Tính 

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)

\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

Câu 3 

a) Ta có : M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹ (1)

=> 3M = 3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰  (2)

Lấy (2) trừ (1) theo vế ta có : 

3M - M = (3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰) - ( M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹)

=>  2M = 3¹²⁰ - 1

=>    M = \(\frac{3^{120}-1}{2}\)

b) M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

        = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

        = (1 + 3 + 3²) + 3³(1 + 3 + 3²) + ... + 3¹¹⁷(1 + 3 + 3²)

        = 13 + 3³.13 + ... + 3¹¹⁷.13

        = 13(1 + 3³ + ... + 3¹¹⁷) \(⋮\)13

=> M \(⋮\)13

M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

= (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + ... + (3¹¹⁶ + 3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

= (1 + 3 + 3² + 3³) + 3⁴(1 + 3 + 3² + 3³) + ... + 3¹¹⁶(1 + 3 + 3² + 3³)

= 40 + 3⁴.40 + ... + 3¹¹⁶.40

= 40(1 + 3⁴ + ... + 3¹¹⁶) 

= 5.8.(1 + 3⁴ + ... + 3¹¹⁶)  \(⋮\)5

4) Tính 

A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1

=> 2A =  2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1

Lấy 2A trừ A theo vế ta có : 

2A - A = (2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1) - (2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1)

=>   A = 2¹⁰¹ - 2¹⁰⁰ - 2¹⁰⁰ + 1

=>   A = 2¹⁰¹ - (2¹⁰⁰ + 2¹⁰⁰) + 1

=>   A  = 2¹⁰¹ - 2¹⁰⁰ . 2 + 1

=>   A = 1

Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

          = 99.100.101 

=> C = 99.100.101 : 3 =  333300

b) Ta có : D = 2² + 4² + 6² + ... + 98²

                    = 2²(1² + 2²  + 3² + ... + 49²

                    =  2² .(1² + 2²  + 3² + ... + 49²)

                    = 2².(1.1 + 2.2 + 3.3 + ... + 49.49)

                    = 2².[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]

                    = 2².[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]

Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50

=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

          = 49.50.51 

=> E = 49.50.51/3 = 41650

Khi đó D = 2².[41650 - (1 + 2 + 3 + 4 + ... + 49)]

               = 2².[41650 - 49(49 + 1)/2]

               = 2².[41650 - 1225 

               = 2².40425

               = 161700

=> D = 161700

\(A=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(A=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(A=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(A=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)

\(A=\frac{2^{18}.3^9.\left(2+5\right)}{2^{18}.3^9.\left(2+2^2.3\right)}\)

\(A=1.\frac{2+5}{2+4.3}\)

\(A=\frac{7}{14}=\frac{1}{2}\)

Vậy \(A=\frac{1}{2}\)

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}=\frac{2^{18}.3^9\left(2+5\right)}{2^{18}.3^9\left(2+3.2^2\right)}=\frac{7}{14}=\frac{1}{2}\)