Phạm Minh Thảo
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2026-01-30 20:54:35
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2026-01-30 20:47:25
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2026-01-30 19:11:13
Bước 1: Giải câu a Gọi d=UCLN(2n+1,-3n−1)d equals cap U cap C cap L cap N open paren 2 n plus 1 comma negative 3 n minus 1 close paren𝑑=𝑈𝐶𝐿𝑁(2𝑛+1,−3𝑛−1). Ta có:
Bài 2: So sánh dãy số Bước 1: Giải câu a Ta có 1n2<1(n−1)nthe fraction with numerator 1 and denominator n squared end-fraction is less than the fraction with numerator 1 and denominator open paren n minus 1 close paren n end-fraction1𝑛2<1(𝑛−1)𝑛. Áp dụng:
A=132+142+...+11002<12.3+13.4+...+199.100cap A equals the fraction with numerator 1 and denominator 3 squared end-fraction plus the fraction with numerator 1 and denominator 4 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 100 squared end-fraction is less than 1 over 2.3 end-fraction plus 1 over 3.4 end-fraction plus point point point plus 1 over 99.100 end-fraction𝐴=132+142+...+11002<12.3+13.4+...+199.100
A<12−13+13−14+...+199−1100=12−1100<12cap A is less than one-half minus one-third plus one-third minus one-fourth plus point point point plus 1 over 99 end-fraction minus 1 over 100 end-fraction equals one-half minus 1 over 100 end-fraction is less than one-half𝐴<12−13+13−14+...+199−1100=12−1100<12. Bước 2: Giải câu b A=122+132+...+120142<11.2+12.3+...+12013.2014cap A equals the fraction with numerator 1 and denominator 2 squared end-fraction plus the fraction with numerator 1 and denominator 3 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 2014 squared end-fraction is less than 1 over 1.2 end-fraction plus 1 over 2.3 end-fraction plus point point point plus 1 over 2013.2014 end-fraction𝐴=122+132+...+120142<11.2+12.3+...+12013.2014
A<1−12014<1cap A is less than 1 minus 1 over 2014 end-fraction is less than 1𝐴<1−12014<1. Tuy nhiên, để chứng minh A<34cap A is less than three-fourths𝐴<34, ta giữ nguyên số hạng đầu:
A=14+(132+...+120142)<14+(12.3+...+12013.2014)=14+(12−12014)=34−12014<34cap A equals one-fourth plus open paren the fraction with numerator 1 and denominator 3 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 2014 squared end-fraction close paren is less than one-fourth plus open paren 1 over 2.3 end-fraction plus point point point plus 1 over 2013.2014 end-fraction close paren equals one-fourth plus open paren one-half minus 1 over 2014 end-fraction close paren equals three-fourths minus 1 over 2014 end-fraction is less than three-fourths𝐴=14+(132+...+120142)<14+(12.3+...+12013.2014)=14+(12−12014)=34−12014<34.
Bài 3: Tìm n và x, y Bước 1: Giải câu a A=2n+22n−4=2n−4+62n−4=1+62n−4=1+3n−2cap A equals the fraction with numerator 2 n plus 2 and denominator 2 n minus 4 end-fraction equals the fraction with numerator 2 n minus 4 plus 6 and denominator 2 n minus 4 end-fraction equals 1 plus the fraction with numerator 6 and denominator 2 n minus 4 end-fraction equals 1 plus the fraction with numerator 3 and denominator n minus 2 end-fraction𝐴=2𝑛+22𝑛−4=2𝑛−4+62𝑛−4=1+62𝑛−4=1+3𝑛−2.
Để Acap A𝐴 là số tự nhiên, 3n−2the fraction with numerator 3 and denominator n minus 2 end-fraction3𝑛−2phải là số nguyên và 1+3n−2≥01 plus the fraction with numerator 3 and denominator n minus 2 end-fraction is greater than or equal to 01+3𝑛−2≥0.
⇒n−2∈{1,3,-1,-3}⇒n∈{3,5,1,-1}implies n minus 2 is an element of the set 1 comma 3 comma negative 1 comma negative 3 end-set implies n is an element of the set 3 comma 5 comma 1 comma negative 1 end-set⇒𝑛−2∈{1,3,−1,−3}⇒𝑛∈{3,5,1,−1}.
Kiểm tra điều kiện A∈Ncap A is an element of the natural numbers𝐴∈ℕ:
Vì x,y∈Zx comma y is an element of the integers𝑥,𝑦∈ℤ, ta có các cặp:
Bài 5: Dãy số quy luật Bước 1: Biến đổi vế trái Ta có công thức: 1n(n+2)=12(1n−1n+2)the fraction with numerator 1 and denominator n open paren n plus 2 close paren end-fraction equals one-half open paren 1 over n end-fraction minus the fraction with numerator 1 and denominator n plus 2 end-fraction close paren1𝑛(𝑛+2)=12(1𝑛−1𝑛+2).
Vế trái (VT) trở thành:
VT=12(1−13+13−15+...+1x−1x+2)=12(1−1x+2)cap V cap T equals one-half open paren 1 minus one-third plus one-third minus one-fifth plus point point point plus 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren equals one-half open paren 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren𝑉𝑇=12(1−13+13−15+...+1𝑥−1𝑥+2)=12(1−1𝑥+2) Bước 2: Tìm x 12(1−1x+2)=512⇒1−1x+2=56⇒1x+2=16one-half open paren 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren equals 5 over 12 end-fraction implies 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction equals five-sixths implies the fraction with numerator 1 and denominator x plus 2 end-fraction equals one-sixth12(1−1𝑥+2)=512⇒1−1𝑥+2=56⇒1𝑥+2=16
⇒x+2=6⇒x=4implies x plus 2 equals 6 implies x equals 4⇒𝑥+2=6⇒𝑥=4.
- (2n+1)⋮d⇒3(2n+1)⋮d⇒(6n+3)⋮dopen paren 2 n plus 1 close paren ⋮ d implies 3 open paren 2 n plus 1 close paren ⋮ d implies open paren 6 n plus 3 close paren ⋮ d(2𝑛+1)⋮𝑑⇒3(2𝑛+1)⋮𝑑⇒(6𝑛+3)⋮𝑑
- (-3n−1)⋮d⇒2(-3n−1)⋮d⇒(-6n−2)⋮dopen paren negative 3 n minus 1 close paren ⋮ d implies 2 open paren negative 3 n minus 1 close paren ⋮ d implies open paren negative 6 n minus 2 close paren ⋮ d(−3𝑛−1)⋮𝑑⇒2(−3𝑛−1)⋮𝑑⇒(−6𝑛−2)⋮𝑑
Cộng hai biểu thức: [(6n+3)+(-6n−2)]⋮d⇒1⋮d⇒d=1open bracket open paren 6 n plus 3 close paren plus open paren negative 6 n minus 2 close paren close bracket ⋮ d implies 1 ⋮ d implies d equals 1[(6𝑛+3)+(−6𝑛−2)]⋮𝑑⇒1⋮𝑑⇒𝑑=1.
Vậy Acap A𝐴 là phân số tối giản.
- (4n−3)⋮d⇒5(4n−3)⋮d⇒(20n−15)⋮dopen paren 4 n minus 3 close paren ⋮ d implies 5 open paren 4 n minus 3 close paren ⋮ d implies open paren 20 n minus 15 close paren ⋮ d(4𝑛−3)⋮𝑑⇒5(4𝑛−3)⋮𝑑⇒(20𝑛−15)⋮𝑑
- (5n−4)⋮d⇒4(5n−4)⋮d⇒(20n−16)⋮dopen paren 5 n minus 4 close paren ⋮ d implies 4 open paren 5 n minus 4 close paren ⋮ d implies open paren 20 n minus 16 close paren ⋮ d(5𝑛−4)⋮𝑑⇒4(5𝑛−4)⋮𝑑⇒(20𝑛−16)⋮𝑑
Trừ hai biểu thức: [(20n−15)−(20n−16)]⋮d⇒1⋮d⇒d=1open bracket open paren 20 n minus 15 close paren minus open paren 20 n minus 16 close paren close bracket ⋮ d implies 1 ⋮ d implies d equals 1[(20𝑛−15)−(20𝑛−16)]⋮𝑑⇒1⋮𝑑⇒𝑑=1.
Vậy Bcap B𝐵 là phân số tối giản.
Bài 2: So sánh dãy số Bước 1: Giải câu a Ta có 1n2<1(n−1)nthe fraction with numerator 1 and denominator n squared end-fraction is less than the fraction with numerator 1 and denominator open paren n minus 1 close paren n end-fraction1𝑛2<1(𝑛−1)𝑛. Áp dụng:
A=132+142+...+11002<12.3+13.4+...+199.100cap A equals the fraction with numerator 1 and denominator 3 squared end-fraction plus the fraction with numerator 1 and denominator 4 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 100 squared end-fraction is less than 1 over 2.3 end-fraction plus 1 over 3.4 end-fraction plus point point point plus 1 over 99.100 end-fraction𝐴=132+142+...+11002<12.3+13.4+...+199.100
A<12−13+13−14+...+199−1100=12−1100<12cap A is less than one-half minus one-third plus one-third minus one-fourth plus point point point plus 1 over 99 end-fraction minus 1 over 100 end-fraction equals one-half minus 1 over 100 end-fraction is less than one-half𝐴<12−13+13−14+...+199−1100=12−1100<12. Bước 2: Giải câu b A=122+132+...+120142<11.2+12.3+...+12013.2014cap A equals the fraction with numerator 1 and denominator 2 squared end-fraction plus the fraction with numerator 1 and denominator 3 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 2014 squared end-fraction is less than 1 over 1.2 end-fraction plus 1 over 2.3 end-fraction plus point point point plus 1 over 2013.2014 end-fraction𝐴=122+132+...+120142<11.2+12.3+...+12013.2014
A<1−12014<1cap A is less than 1 minus 1 over 2014 end-fraction is less than 1𝐴<1−12014<1. Tuy nhiên, để chứng minh A<34cap A is less than three-fourths𝐴<34, ta giữ nguyên số hạng đầu:
A=14+(132+...+120142)<14+(12.3+...+12013.2014)=14+(12−12014)=34−12014<34cap A equals one-fourth plus open paren the fraction with numerator 1 and denominator 3 squared end-fraction plus point point point plus the fraction with numerator 1 and denominator 2014 squared end-fraction close paren is less than one-fourth plus open paren 1 over 2.3 end-fraction plus point point point plus 1 over 2013.2014 end-fraction close paren equals one-fourth plus open paren one-half minus 1 over 2014 end-fraction close paren equals three-fourths minus 1 over 2014 end-fraction is less than three-fourths𝐴=14+(132+...+120142)<14+(12.3+...+12013.2014)=14+(12−12014)=34−12014<34.
Bài 3: Tìm n và x, y Bước 1: Giải câu a A=2n+22n−4=2n−4+62n−4=1+62n−4=1+3n−2cap A equals the fraction with numerator 2 n plus 2 and denominator 2 n minus 4 end-fraction equals the fraction with numerator 2 n minus 4 plus 6 and denominator 2 n minus 4 end-fraction equals 1 plus the fraction with numerator 6 and denominator 2 n minus 4 end-fraction equals 1 plus the fraction with numerator 3 and denominator n minus 2 end-fraction𝐴=2𝑛+22𝑛−4=2𝑛−4+62𝑛−4=1+62𝑛−4=1+3𝑛−2.
Để Acap A𝐴 là số tự nhiên, 3n−2the fraction with numerator 3 and denominator n minus 2 end-fraction3𝑛−2phải là số nguyên và 1+3n−2≥01 plus the fraction with numerator 3 and denominator n minus 2 end-fraction is greater than or equal to 01+3𝑛−2≥0.
⇒n−2∈{1,3,-1,-3}⇒n∈{3,5,1,-1}implies n minus 2 is an element of the set 1 comma 3 comma negative 1 comma negative 3 end-set implies n is an element of the set 3 comma 5 comma 1 comma negative 1 end-set⇒𝑛−2∈{1,3,−1,−3}⇒𝑛∈{3,5,1,−1}.
Kiểm tra điều kiện A∈Ncap A is an element of the natural numbers𝐴∈ℕ:
- n=3⇒A=4n equals 3 implies cap A equals 4𝑛=3⇒𝐴=4 (TM)
- n=5⇒A=2n equals 5 implies cap A equals 2𝑛=5⇒𝐴=2 (TM)
- n=1⇒A=-2n equals 1 implies cap A equals negative 2𝑛=1⇒𝐴=−2 (Loại)
- n=-1⇒A=0n equals negative 1 implies cap A equals 0𝑛=−1⇒𝐴=0 (TM)
Đáp số: n∈{3,5,-1}n is an element of the set 3 comma 5 comma negative 1 end-set𝑛∈{3,5,−1}.
Vì x,y∈Zx comma y is an element of the integers𝑥,𝑦∈ℤ, ta có các cặp:
- x−1=1,y−2=3⇒x=2,y=5x minus 1 equals 1 comma y minus 2 equals 3 implies x equals 2 comma y equals 5𝑥−1=1,𝑦−2=3⇒𝑥=2,𝑦=5
- x−1=3,y−2=1⇒x=4,y=3x minus 1 equals 3 comma y minus 2 equals 1 implies x equals 4 comma y equals 3𝑥−1=3,𝑦−2=1⇒𝑥=4,𝑦=3
- x−1=-1,y−2=-3⇒x=0,y=-1x minus 1 equals negative 1 comma y minus 2 equals negative 3 implies x equals 0 comma y equals negative 1𝑥−1=−1,𝑦−2=−3⇒𝑥=0,𝑦=−1
- x−1=-3,y−2=-1⇒x=-2,y=1x minus 1 equals negative 3 comma y minus 2 equals negative 1 implies x equals negative 2 comma y equals 1𝑥−1=−3,𝑦−2=−1⇒𝑥=−2,𝑦=1
Bài 5: Dãy số quy luật Bước 1: Biến đổi vế trái Ta có công thức: 1n(n+2)=12(1n−1n+2)the fraction with numerator 1 and denominator n open paren n plus 2 close paren end-fraction equals one-half open paren 1 over n end-fraction minus the fraction with numerator 1 and denominator n plus 2 end-fraction close paren1𝑛(𝑛+2)=12(1𝑛−1𝑛+2).
Vế trái (VT) trở thành:
VT=12(1−13+13−15+...+1x−1x+2)=12(1−1x+2)cap V cap T equals one-half open paren 1 minus one-third plus one-third minus one-fifth plus point point point plus 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren equals one-half open paren 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren𝑉𝑇=12(1−13+13−15+...+1𝑥−1𝑥+2)=12(1−1𝑥+2) Bước 2: Tìm x 12(1−1x+2)=512⇒1−1x+2=56⇒1x+2=16one-half open paren 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction close paren equals 5 over 12 end-fraction implies 1 minus the fraction with numerator 1 and denominator x plus 2 end-fraction equals five-sixths implies the fraction with numerator 1 and denominator x plus 2 end-fraction equals one-sixth12(1−1𝑥+2)=512⇒1−1𝑥+2=56⇒1𝑥+2=16
⇒x+2=6⇒x=4implies x plus 2 equals 6 implies x equals 4⇒𝑥+2=6⇒𝑥=4.
2026-01-30 19:04:29
335802456
2026-01-30 19:03:51
1 Yêu Tổ quốc, yêu đồng bào
2 Học tập tốt, lao động tốt
3 Đoàn kết tốt, kỷ luật tốt
4 Giữ gìn vệ sinh thật tốt
5 Khiêm tốn, thật thà, dũng cảm
2026-01-29 21:44:37
Vây mình chúc bạn vui vẻ
2026-01-29 21:42:43
Rồi sao
2026-01-29 21:40:38
Lười
2026-01-29 21:38:56
Ok , chỉ là bạn đăng hơi lộ liễu với những người thích nói câu ai hỏi thôi 😶
2026-01-29 21:15:21
còn