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a, \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)
b, \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+...+\dfrac{5}{99.101}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)
Vậy...
2.a) n+7/n+3 là số nguyên
=>n+7¤n+3(¤ là chia hết cho)
(n+3)+4¤n+3
=>4¤n+3
=>n+3€Ư(4)={1;-1;2;-2;4;-4} (€ là thuộc)
+)n+3=1
n=1-3
n=-2
+)n+3=-1
n=-1-3
n=-4
+)n+3=2
n=2-3
n=-1
+)n+3=-2
n=-2-3
n=-5
+)n+3=4
n=4-3
n=1
+)n+3=-4
n=-4-3
n=-7
Vậy n€{-2;-4;-5;-7;+-1)
b)3/1.3....3/2017.2019
=3/2.(2/1.3...2/2017.2019)
=3/2.[1-(1/3)+(1/3)-(1/5)+...+(1/2017)-(1/2019)]
=3/2.[1-(1/2019)]
=3/2.2018/2019
=1009/673
1.a)1/3+3/8《x/24<5/24+5/8(《 là < hoặc =)
17/24《x/24<20/24
=>x/24€{17/24;18/24;19/24}
=>x€{17;18;19}
b)Chỗ đất đã dùng là
3/10+7/10=10/10=1
Vậy thì làm gì còn đất nữa mà nói: "phần đất còn lại..." làm gì cho bất công
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
Trả lời
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)
=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(2.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(2.\dfrac{100}{101}\)
=\(\dfrac{200}{101}\)
Cho \(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{1}{7}.\dfrac{1}{2}+\dfrac{15}{8}}{a+\dfrac{5}{6}-\left(\dfrac{-1}{3}\right)}\)
a) Rút gọn A?
b) Tính A khi a=75%
c) Tìm a để A=50%
d) Tìm a thuộc Z để A là số nguyên.
e) Với a = bao nhiêu để A có giá trị bằng với giá trị của biểu thức:
\(B=\dfrac{\dfrac{2}{3}.\dfrac{15}{6}+\left(-0,5\right)^3}{\dfrac{1}{9}.6^2-5\dfrac{1}{3}}\)
Giải
a, Ta có:
\(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{3}{7}.\dfrac{1}{6}+\dfrac{1}{8}.15}{a+\dfrac{5}{6}+\dfrac{1}{3}}\)
\(A=\dfrac{\dfrac{3}{7}.\left(\dfrac{-5}{8}+\dfrac{3}{4}+\dfrac{1}{6}\right)+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{3}{7}.\dfrac{7}{24}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}.\left(15+1\right)}{a+\dfrac{7}{6}}\)
\(A=\dfrac{2}{a+\dfrac{7}{6}}\)
b, Thay \(a=75\%\) vào \(A\), ta được:
\(A=\dfrac{2}{75\%+\dfrac{7}{6}}\)
\(A=\dfrac{2}{\dfrac{3}{4}+\dfrac{7}{6}}\)
\(\Rightarrow A=\dfrac{23}{12}\)
c, Ta có: \(\dfrac{2}{a+\dfrac{7}{6}}=50\%\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{1}{2}\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{2}{4}\)
\(\Rightarrow a+\dfrac{7}{6}=4\)
\(\Rightarrow a=\dfrac{17}{6}\)
d, Để \(A\in Z\Rightarrow2⋮a+\dfrac{7}{6}\)
\(\Rightarrow a+\dfrac{7}{6}\in\left\{\pm1;\pm2\right\}\)
\(\circledast,a+\dfrac{7}{6}=1\Rightarrow a=\dfrac{-1}{6}\)
\(\circledast,a+\dfrac{7}{6}=-1\Rightarrow a=\dfrac{-13}{6}\)
\(\circledast,a+\dfrac{7}{6}=2+\Rightarrow a=\dfrac{5}{6}\)
\(\circledast,a+\dfrac{7}{6}=-2\Rightarrow a=\dfrac{-19}{6}\)
\(a\in\varnothing\) khi \(A\in Z\)
e, Ta có:
\(B=\dfrac{5}{3}+\dfrac{-1}{8}\Rightarrow B=\dfrac{37}{24}\)
\(\Rightarrow\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{37}{24}\)
\(a+\dfrac{7}{6}=\dfrac{37}{24}.2\)
\(a+\dfrac{7}{6}=\dfrac{37}{12}\)
\(\Rightarrow a=\dfrac{23}{12}\)
Chúc bạn học thiệt giỏi nha!!! 
)
1/
a/ A = 1 + 3 + 3^2 + 3^3 + ... + 3^119
=> 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^120
=> 3A - A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^120 - (1 + 3 + 3^2 + 3^3 + ... + 3^119)
=> 2A = 3^120 - 1
=> A = (3 ^120 - 1)/2
b/ 2A + 1 = 27x
<=> 3^120 = 27x
<=> 27^40 = 27x
<=> x = 40
c/ +) A = 1 + 3 + 3^2 + 3^3 + ... + 3^119
= (1 + 3^2) + (3 + 3^3) + (3^4 + 3^6) + ...+ (3^117 + 3^119)
= 1+ 3^2 + 3(1+ 3^2) + 3^4(1 + 3^2) ...+ 3^117( 1+ 3^2)
= (1 + 3^2) (1 + 3 + 3^4+ ...+ 3^117)
= 10 * (1 + 3 + 3^4+ ...+ 3^117) \(⋮\) 5
+) A = 1 + 3 + 3^2 + 3^3 + ... + 3^119
= (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ...+ (3^117 + 3^118 + 3^119)
= (1 + 3 + 3^2) + 3^3 (1+ 3 + 3^2) + ...+ 3^117 (1+ 3 + 3^2)
= (1 + 3 + 3^2) (1+ 3^3 +... + 3^117)
= 13 * (1+ 3^3 +... + 3^117) \(⋮\)13
\(x+8-(x+22)=x+8-x-22=8-22=-14\)
\(-(x+5)+(x+10)-5=-x-5+x+10-5=0\)
1*5* \(⋮\)2;3;5;6;9
Vì 1*5* chia hết cho 2 và 5 nên dấu sao cuối cùng=0
Ta có: 1*5* chia hết cho 6=> chia hết cho 3 và 2
1*5* chia hết cho 9
1*50 chia hết cho 9
1+*+5+0 chia hết cho 9
6+* chia hết cho 9=> *=3
vậy số cần tìm là 1350
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+.....+\dfrac{5}{99.101}\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)
Để \(A\in Z\)thì
\(n+2⋮n-5\)
\(n-5+7⋮n-5\)
\(\Leftrightarrow7⋮n-5\)
\(\Leftrightarrow n-5\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(n\in\left\{6;4;12;-2\right\}\)
1 bỏ dấu ngoặc rồi tính :
a) x+ 8 - ( x + 22)
= x + 8 - x - 22
= -14
b) -(x+5) + (x + 10 ) - 5
= -x - 5 + x + 10 -5
= 0