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(101+102+...+200)+(-1-2-3-...-100)
=(101-1)+(102-2)+...+(200-100)
=100+100+...+100
=100*100=10000
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(0,75+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{3}{4}+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{5}{12}-\dfrac{5}{18}=\dfrac{5}{36}\)
c) \(\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{15}{20}=\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{5}{45}\cdot\dfrac{3}{4}=\dfrac{15}{180}=\dfrac{1}{12}\)
d) \(\left(\dfrac{-1}{9}\right)\cdot\left(\dfrac{15}{22}\right):\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}:\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}\cdot\left(\dfrac{9}{-25}\right)=\dfrac{-3}{-110}=\dfrac{3}{110}\)
a) \(0,75\) + \(\dfrac{-1}{3}\) - \(\dfrac{5}{18}\)= \(\dfrac{5}{12}\) - \(\dfrac{5}{18}\) = \(\dfrac{5}{36}\)
b) \(\dfrac{4}{15}\)x \(\dfrac{1}{3}\)x \(\dfrac{15}{20}\)= 4/45 x 15/20 = 1/15
c) -1/9 x 15/22 : -25/9 = -5/66 : -25/9 = 3/110
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Bài 1:
\(A=3+3^2+...+3^{100}\)
=>\(3\cdot A=3^2+3^3+...+3^{101}\)
=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2A=3^{101}-3\)
=>\(2A+3=3^{101}\)
mà \(2A+3=3^n\)
nên n=101
Bài 2:
a: \(M=3+3^2+3^3+3^4+...+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{98}\right)⋮12\)
=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)
b: \(M=3+3^2+...+3^{100}\)
=>\(3M=3^2+3^3+...+3^{101}\)
=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2M=3^{101}-3\)
=>\(2M+3=3^{101}\)
=>n=101
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a) 4x - 15 = -75 -x
4x+x = -75 + 15
5x = 60
x= 60: 5
=> x= 12
b) 3| x-7| = 21
|x-7|= 21:3
|x-7|=7
=> x-7 =7 hoặc x-7=-7
+) x-7=7
x=7+7=14
+) x-7=-7
x= -7+7=0
=> x=14 hoặc x=0
c) Áp dụng t/c phân số bằng nhau
=> x= \(\frac{-3.\left(-2\right)}{6}\)=\(\frac{6}{6}\)=1
Thay x=1 => y= \(\frac{\left(-2\right).\left(-18\right)}{1}\)=\(\frac{36}{1}\)=36
Thay y =36 => z=\(\frac{\left(-18\right).24}{36}\)=\(\frac{-432}{36}\)=-12
vậy (x,y,z)= (1;36;-12)
(câu d dài quá vs lại cx dễ nên bn tự lm nha mk chỉ giúp đến đây thôi)
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Bài 3:
a: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
\(=\dfrac{11\cdot3^{29}-3^{30}}{3^{28}\cdot2^2}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot4}\)
\(=3\cdot\dfrac{8}{4}=3\cdot2=6\)
b: \(\dfrac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)
\(=\dfrac{2^{10}\cdot3^9\cdot3-2^{10}\cdot3^9\cdot1}{2^9\cdot3^{10}}\)
\(=\dfrac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}\)
\(=\dfrac{2}{3}\cdot2=\dfrac{4}{3}\)
c: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
Bài 4:
a: \(\dfrac{10^2+11^2+12^2}{13^2+14^2}\)
\(=\dfrac{100+121+144}{169+195}\)
\(=\dfrac{365}{365}=1\)
b: \(\dfrac{2^3\cdot9^4+9^3\cdot45}{9^2\cdot10-9^2}\)
\(=\dfrac{2^3\cdot9^4+9^3\cdot9\cdot5}{9^2\left(10-1\right)}\)
\(=\dfrac{9^4\left(2^3+5\right)}{9^3}=9\cdot\left(8+5\right)=9\cdot13=117\)
c: \(\left[\dfrac{3^{14}\cdot69+3^{14}\cdot12}{3^{16}}-7\right]:2^4\)
\(=\left(\dfrac{3^{14}\left(69+12\right)}{3^{14}\cdot9}-7\right):16\)
\(=\dfrac{\left(\dfrac{81}{9}-7\right)}{16}=\dfrac{2}{16}=\dfrac{1}{8}\)
d: \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(\dfrac{24}{3}\right)^4-\left(\dfrac{32}{16}\right)^{12}\)
\(=8^4-2^{12}\)
\(=2^{12}-2^{12}=0\)
e: \(2010^{2010}\left(7^{10}:7^8-3\cdot2^4-2^{2010}:2^{2010}\right)\)
\(=2010^{2010}\left(7^2-3\cdot16-1\right)\)
\(=2010^{2010}\cdot\left(49-48-1\right)\)
=0
f: \(\dfrac{2^{100}+2^{101}+2^{102}}{2^{97}+2^{98}+2^{99}}\)
\(=\dfrac{2^{100}\left(1+2+2^2\right)}{2^{97}\left(1+2+2^2\right)}\)
\(=\dfrac{2^{100}}{2^{97}}=2^3=8\)
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\(\left(x+2,5\right):2\frac{1}{2}-1,5=-1\frac{1}{4}\\ \left(x+2,5\right):2,5=-1\frac{1}{4}+1,5\\ x:2,5+1=\frac{-1}{4}\\ x:\frac{5}{2}=\frac{-5}{4}\\ x=\frac{-1}{2}\)
\(\dfrac{1}{8}-\dfrac{4}{8}=\dfrac{-3}{8}\)
-3/8