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a) (-37) + 14 + 26 + 37
= [(-37) + 37] + (14 + 26)
= 0 + 40 = 40
b) (-24) + 6 + 10 + 24
= [(-24) + 24] + (10 + 6)
= 0 + 16 = 16
c) 15 + 23 + (-25) + (-23)
= [15 + (-25)] + [23 + (-23)]
= (-10) + 0 = -10
d) 60 + 33 + (-50) + (-33)
= [60 + (-50)] + [33 + (-33)]
= 10 + 0 = 10
e) (-16) + (-209) + (-14) + 209
= [(-16) + (-14)] + [(-209) + 209]
= (-30) + 0 = -30
f) \(-3^2+\left(-54\right)\div\left[\left(-2\right)^8+7\right]\times\left(-2\right)^2\\ =\left(-9\right)+\left(-54\right)\div263\times4\\ =\left(-9\right)+\dfrac{-216}{263}=\dfrac{-2583}{263}\)
a. \(\left[\left(-37\right)+37\right]+\left(14+16\right)\) = 30
B. \(\left[\left(-24\right)+24\right]+\left(10+6\right)\) = 16
C. \(\left[\left(-23\right)+23\right]+\left(15-23\right)\)= -8
d. \(\left[33-33\right]+\left(60-50\right)\) = 10
e. \(\left(209-209\right)+\left(-16-14\right)\)= -30
Lời giải:
Đặt biểu thức là $A$
\(A=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+....+\frac{1}{3^{99}}+\frac{1}{3^{101}}\)
\(3^2.A=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
Trừ theo vế:
\(8A=3-\frac{1}{3^{101}}\Rightarrow A=\frac{3}{8}-\frac{1}{8.3^{101}}\)
Akai Haruma Giáo viên Giúp em câu em gửi trong inb nhé chị
P/s : Sorry bạn chủ tus nhé , mình lượn ngay đây
\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)
S = 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39
S = ( 3 + 32 + 33 ) +34 + 35 + 36 + 37 + 38 + 39
S = 39 + 34 + 35 + 36 + 37 + 38 + 39
Vì 39 ⋮ -39
<=> S ⋮ -39
Ta có \(\frac{33}{-37}=\frac{-33}{37}\)
Mà \(\frac{-33}{37}>\frac{-34}{37}>\frac{-34}{35}\)=>\(\frac{-33}{37}>\frac{-34}{35}\)
Vậy \(\frac{33}{-37}>\frac{-34}{35}\)
Vì \(\frac{2019}{2020}=1-\frac{2019}{2020}=\frac{1}{2020}\)
\(\frac{2006}{2007}=1-\frac{2006}{2007}=\frac{1}{2007}\)
Vì \(\frac{1}{2020}< \frac{1}{2007}=>\frac{2019}{2020}>\frac{2006}{2007}\)
Vậy\(\frac{2019}{2020}>\frac{2006}{2007}\)
Chiều nay mình giải tiếp cho.
Bài 1:
1) \(9A=3^3+3^5+...+3^{113}\)
\(\Rightarrow8A=9A-A=3^3+3^5+...+3^{113}-3-3^3-...-3^{111}=3^{113}-3\)
\(\Rightarrow A=\dfrac{3^{113}-3}{8}\)
2) \(9B=3^4+3^6+...+3^{202}\)
\(\Rightarrow8B=9B-B=3^4+3^6+...+3^{202}-3^2-3^4-...-3^{200}=3^{202}-3^2=3^{202}-9\)
\(\Rightarrow B=\dfrac{3^{202}-9}{8}\)
3) \(25C=5^3+5^5+...+5^{101}\)
\(\Rightarrow24C=25C-C=5^3+5^5+...+5^{101}-5-5^3-...-5^{99}=5^{101}-5\)
\(\Rightarrow C=\dfrac{5^{101}-5}{24}\)
4) \(25D=5^4+5^6+...+5^{102}\)
\(\Rightarrow24D=25D-D=5^4+5^6+...+5^{102}-5^2-5^4-...-5^{100}=5^{102}-25\)
\(\Rightarrow D=\dfrac{5^{102}-25}{24}\)
Bài 2:
a) Gọi d là UCLN(2n+1,n+1)
\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\n+1⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\2n+2⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\Rightarrow1⋮d\)
Vậy 2n+1 và n+1 là 2 số nguyên tố cùng nhau
\(\Rightarrow\dfrac{2n+1}{n+1}\) là phân số tối giản
b) Gọi d là UCLN(2n+3,3n+4)
\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+8⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\Rightarrow1⋮d\)
\(\Rightarrow\dfrac{2n+3}{3n+4}\) là phân số tối giản
\(\frac{20\cdot3^{37}+2^{35}\cdot45}{5\cdot3^{37}+45\cdot2^{33}}\)
\(=\frac{2^2\cdot5\cdot3^{37}+2^{35}\cdot5\cdot3^2}{5\cdot3^{37}+5\cdot3^2\cdot2^{33}}\)
\(=\frac{2^2\cdot5\cdot3^2\cdot\left(3^{35}+2^{33}\right)}{5\cdot3^2\left(3^{35}+2^{33}\right)}\)
\(=2^2=4\)
\(\frac{-33}{37}>\frac{-34}{37}>\frac{-34}{35}\)
suy ra\(\frac{-33}{37}>\frac{-34}{35}\)
S = 1 + 3 + 32 + 33 +... + 32014
3S = 3 + 32 + 33 + 34 + ... + 32015
3S - S = ( 3 + 32 + 33 + 34 + ... + 32015) - (1 + 3 + 32 + 33 +... + 32014)
2S = 32015 - 1
S = \(\dfrac{3^{2015}-1}{2}\)
Ta có 33/37<33/35
=> -33/37>-33/35
Lại 33/35<34/35
=> -33/35>-34/35
=> -33/37>-34/35
Đặt \(A=3^3+3^5+...+3^{1001}\)
=>\(9A=3^5+3^7+...+3^{1003}\)
=>\(9A-A=3^5+3^7+...+3^{1003}-3^3-3^5-...-3^{1001}\)
=>\(8A=3^{1003}-27\)
=>\(A=\dfrac{3^{1003}-27}{8}\)
\(S=1+3^3+3^5+...+3^{1001}\)
\(=1+\dfrac{3^{1003}-27}{8}=\dfrac{3^{1003}-19}{8}\)
\(S=1+3^3+3^5+...+3^{1001}\)
\(9S=9+3^5+3^7+...+3^{1003}\)
\(9S-S=3^{1003}+9-\left(1+3^3\right)\)
\(8S=3^{1003}-19\)
\(S=\dfrac{3^{1003}-19}{8}\)