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\(210-\left(2^3.3^3+54\right):3^3-50\)
\(=210-270:27-50\)
\(=210-10-50\)
\(=200-50=150\)
\(\left[210-\left(2^3.3^3+54\right):3^3-50\right]\)
\(=210-270:27-50\)
\(=210-10-50\)
\(=200-50\)
\(=150\)
a) 450 : ( 50 - 7x ) = 30
=> 50 - 7x = 15
=> 7x = 35
=> x =5
Vậy x=5
b) 14x - 2x = 120
=> x ( 14 - 2 ) = 120
=> x . 12 = 120
=> x =10
Vậy x = 10
c) x.13 - x.10 = 240
=> x ( 13 - 10 ) = 240
=> x. 3 = 240
=> x = 80
Vậy x = 80
d) 28x - 17x - x = 250
=> x( 28 - 17 - 1 ) = 250
=> x. 10 = 250
=> x = 25
Vậy x = 25
e) 34x - 14x = 200
=> x ( 34 - 14 ) = 200
=> x.20 = 200
=> x = 10
Vậy x = 10
f) 280 - x.9 - x =80
=> x.9 - x = 200
=> x ( 9 -1 ) = 200
=> x.8 = 200
=> x = 25
Vậy x = 25
a) 450 : (50-7.x) = 30
Ta có: 50-7.x = y
=> 450 : y = 30
y = 15
=> 50-7. x= 15
7x = 35
x= 5
vậy x = 5
(bạn ơi câu này mình trả lời đúng nhưng cách làm thì hơi khác 1 tí)
b) x.14 - x.2 = 120
x . (14-2) = 120
x . 12 = 120
x = 120 : 12
x = 10
Vậy x= 10
c) x.13 - x.10 = 240
x . (13-10) = 240
x . 3 = 240
x = 80
Vậy x = 80
d) 28.x - x.17-x = 250
28.x - x.17 - x.1 = 250
=> x. (28-17-1) = 250
x. 10 = 250
x = 25
Vậy x=25
e) x.34-x.14 = 200
x. ( 34-14) = 200
x. 20 = 200
x = 10
vậy x = 10
f) 280 - x.9 - x = 80
280 - x.9 - x.1 = 80
280 - x.( 9-1) = 80
280 - x.8 = 80
x.8 = 200
x = 25
vậy x = 25
a) => x : 80 = 2
=> x = 160
b) => 200/15 : x = 3
=> x = 40/9
c) => 158 - x = 140
=> x = 18
d) => 231 - (x-6) = 103
=> x - 6 = 128
=> x = 134
\(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+....+\frac{1}{240}=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{15.16}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{7}-\frac{1}{16}=\frac{9}{112}\)
\(=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{7}-\frac{1}{16}\)
\(=\frac{9}{112}\)
\(E=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow E=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)
\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2.\frac{3}{16}=\frac{3}{8}\)
\(E=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(E=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(E=2.\frac{3}{16}\)
\(E=\frac{3}{8}\)
\(B-\left(\frac{1}{30.33}+\frac{1}{33.36}+...+\frac{1}{117.120}\right)=\frac{1}{120}-\frac{3}{30.33}-\frac{3}{33.36}-...-\frac{3}{117.120}\)
\(B-\frac{1}{3}\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-...+\frac{1}{117}-\frac{1}{120}\right)=\frac{1}{120}-\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-...-\frac{1}{120}\right)\)
\(\Rightarrow B-\frac{1}{3}\left(\frac{1}{30}-\frac{1}{120}\right)=\frac{1}{120}-\frac{1}{30}+\frac{1}{120}\)
\(\Rightarrow B=\frac{1}{60}-\frac{1}{30}+\frac{1}{3}\left(\frac{1}{30}-\frac{1}{120}\right)=-\frac{1}{120}\)
Phan h moi so ra TSNT roi nhan len la ra!!
a,ta co:
\(220=2^2.5.11\)
\(240=2^4.3.5\)
\(300=2.3.5^2\)
\(\Rightarrow UCLN\left(220,240,300\right)=2.3.5=30\)
\(BCNN\left(220,240,300\right)=2^4.3.5^2=1200\)
Cac cau con lai tuong tu nha!!
HTDT
a ) Ta có : \(220=2^2\cdot5\cdot11\)
\(240=2^4\cdot3\cdot5\)
\(300=2^2\cdot3\cdot5^2\)
\(\Rightarrow\:ƯCLN\)\(\left(220;240;300\right)=2^2\cdot5=20\)
\(\Rightarrow BCNN\left(220;240;300\right)=2^4\cdot3\cdot5^2\cdot11=13200\)
b) Ta có : \(40=2^3\cdot5\)
\(75=3\cdot5^2\)
\(105=3\cdot5\cdot7\)
\(\RightarrowƯCLN\left(40;75;105\right)=5\)
\(\Rightarrow BCNN\left(40;75;105\right)=2^3\cdot3\cdot5^2\cdot7=4200\)
c)
\(C1:\) Ta có : \(18=2\cdot3^2\)
\(36=2^2\cdot3^2\)
\(72=2^3\cdot3^2\)
\(\Rightarrow\:\)\(ƯCLN\)\(\left(18;36;72\right)=2\cdot3^2\)\(=18\)
\(\Rightarrow BCNN\left(18;36;72\right)=2^3\cdot3^2=72\)
\(C2:\) Vì \(36⋮18;72⋮18;18⋮18\) nên \(\:ƯCLN\)\(\left(18;36;72\right)=18\)
Vì \(72\) chia hết cho \(18;36;72\) nên \(BCNN\left(18;36;72\right)=72\)
Dãy số có số số hạng là:
(10050 - 10) : 10 + 1 = 1005 (số hạng)
Tổng dãy số là:
(10 + 10050) . 1005 : 2 = 5055150
~Study well~
#Bạch_Dương_Chi#
số các số hạng là :
(10050-10) : 10 +1 = 1005
tổng dãy trên là:
(10050 + 10) x 10050 : 2 = 5055150
đáp số : 5055150