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c)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)
\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có 7 số 1)
\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(7+1-\frac{1}{8}=\frac{63}{8}\)
Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé
Chúc bạn học tốt !!!
a. 11 + 12 + 13 +14+15+16+17+18+19
= ( 11 + 19 ) + ( 12 + 18 ) + ( 13 + 17 ) + ( 14 + 16 ) + 15
= 30 + 30 + 30 + 30 + 15
= 120 + 15
= 132
b . 1+2+3+4+5+................+99+100
Dãy trên có tất cả số số hạng là :
( 100 - 1 ) : 1 + 1 = 100 ( số )
Tổng của dãy số trên là :
( 100 + 1 ) x 100 : 2 = 5050
Phần c và phần d bạn làm như phần b
Công thức tính số số hạng : ( số lớn - số bé ) : khoảng cách + 1
Công thức tính tổng : ( số lớn + số bé ) x số số hạng : 2
Hok tốt
a, 48.84
= (22)8.(23)4
= 216.212
= 228
b, 415.515
= (4.5)15
= 2015
c, 210.15 + 210.85
= 210.(15 + 85)
= 210.100
=210.(2.5)2
= 212.52
d, 33.92
= 33 . (32)2
= 33.34
= 37
e, 512.7 - 511.10
= 511.(5.7 - 10)
= 511.25
=511.52
=513
f, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)
= \(x^{1+2+3+...+100}\)
= \(x^{\left(1+100\right).100:2}\)
= \(x^{5050}\)
Ta có: \(A=1.3+2.4+3.5+4.6+...+99.101+100.102\)
\(A=1.\left(1+2\right)+2.\left(2+2\right)+3.\left(3+2\right)+4.\left(4+2\right)+....+99.\left(99+2\right)+100.\left(100+2\right)\)
\(A=\left(1^2+2^2+3^2+4^2+...+99^2+100^2\right)+\left(2+4+6+8+...+198+200\right)\)Đặt \(B=1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\)
\(\Rightarrow B=\left(1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\right)-2^2.\left(1^2+2^2+3^2+4^2+5^2+....+49^2+50^2\right)\)Tính dãy tổng quát \(C=1^2+2^2+3^2+4^2+5^2+...+n^2\)
\(C=1\left(0+1\right)+2\left(1+1\right)+3.\left(2+1\right)+4.\left(3+1\right)+5\left(4+1\right)+...+n\left[\left(n-1\right)+1\right]\)
\(C=\left[1.2+2.3+3.4+4.5+...+\left(n-1\right).n\right]+\left(1+2+3+4+5+....+n\right)\)
\(C=n.\left(n+1\right).\left[\left(n-1\right):3+1:2\right]=n.\left(n+1\right).\left(2n+1\right):6\)
Áp dụng vào B ta được:
\(B=100.101.201:6-4.50.51.101:6=166650\)
\(\Rightarrow A=166650+\left(200+2\right).100:2\)
\(\Rightarrow A=166650+10100=176750\)
Vậy A = 176750
Chúc bạn học tốt!!
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)
\(\Rightarrow x=\dfrac{11}{25}\)
b) \(x\cdot9,85+x\cdot0,15=0,1\)
\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)
\(\Rightarrow x\cdot10=0,1\)
\(\Rightarrow x=\dfrac{0,1}{10}\)
\(\Rightarrow x=0,01\)
c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)
\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)
\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)
\(\Rightarrow2022x=0\)
\(\Rightarrow x=\dfrac{0}{2022}\)
\(\Rightarrow x=0\)
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)
Ta có :
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)
\(\Rightarrow x+\dfrac{103}{50}=5\)
\(\Rightarrow x=5-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{147}{50}\)