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\(A=3a\cdot\left(2+b\right)-a\cdot\left(1-b\right)-4ab\)
\(=6a+3b-a+ab-4ab=5a+3b-3ab\)
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
Làm khâu rút gọn thôi
\(=\frac{15}{x+2}+\frac{42}{3x+6}\)
\(=\frac{15}{x+2}+\frac{42}{3\left(x+2\right)}\)
\(=\frac{3.15+42}{3\left(x+2\right)}\)
\(=\frac{87}{3\left(x+2\right)}\)
\(=\frac{29}{x+2}\)
Câu b có phải để tử chia hết cho mẫu không nhỉ? Không chắc thôi để ngkh làm
a ) A = 1/15 . 75/ x + 2 + 3/8 . 64/ 3x + 6
A = 1.75 / 15.( x + 2 ) + 3.64/8.( 3x + 6 )
A = 1.5/1.( x + 2 ) + 1.8/1.( x + 2 )
b ) A = 1.5/1.( x + 2 ) + 1.8/1.( x + 2 )
A = 5/ x+ 2 + 8/ x + 2
A = 5 + 8 / x + 2
A = 13/ x + 2
Thay x = 37
A = 13 / 37 + 2
A = 13 / 39
A = 1/3
Câu 2 :
B = 1 + 3 + 3^2 + 3^3 + ... + 3^199
3B = 3 + 3^2 + 3^3 + 3^4 + ... + 3^200
3B - B = ( 3 + 3^2 + 3^3 + 3^4 + ... 3^200 ) - ( 1 + 3 + 3^2 + 3^3 + ... + 3^199 )
2B = 3^200 - 1
B = 3^200 - 1 / 2
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
@@@@@@@@@@
HT
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
gọi
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
VÌ \(\frac{2019}{2020}< 1\Rightarrow A< 1\)
VẬY \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)
1. a) P = 4 - ( x - 2 )32
( x - 2 )32 ≥ 0 ∀ x => - ( x - 2 )32 ≤ 0 ∀ x
=> 4 - ( x - 2 )32 ≤ 4 ∀ x
Dấu bằng xảy ra <=> x - 2 = 0 => x = 2
Vậy PMax = 4 khi x = 2
b) Q = 20 - | 3 - x |
| 3 - x | ≥ 0 ∀ x => - | 3 - x | ≤ 0 ∀ x
=> 20 - | 3 - x | ≤ 20 ∀ x
Dấu bằng xảy ra <=> 3 - x = 0 => x = 3
Vậy QMax = 20 khi x = 3
c) C = \(\frac{5}{\left(x-3\right)^2+1}\)
Để C có GTLN => ( x - 3 )2 + 1 nhỏ nhất dương
=> ( x - 3 )2 + 1 = 1
=> ( x - 3 )2 = 0
=> x - 3 = 0
=> x = 3
=> CMax = \(\frac{5}{\left(3-3\right)^2+1}=\frac{5}{1}=5\)khi x = 3
a, \(A=3a.2.b-a.432b-4ab\)
\(=6ab-432ab-4ab=-430ab\)
b, \(A=-430ab=\left(-430\right).\frac{1}{229}.\frac{1}{433}=\frac{-430}{229.433}\)