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a) \(3.\left(10.x\right)=111\)
\(10.x=37\)
\(x=\dfrac{37}{10}\)
b) \(3.\left(10+x\right)=111\)
\(10+x=37\)
\(x=27\)
c) \(3+\left(10.x\right)=111\)
\(10.x=108\)
\(x=\dfrac{54}{5}\)
d) \(3+\left(10+x\right)=111\)
\(x=111-3-10\)
\(x=98\)
a)
3 . ( 10 . x ) = 111
3 . 10 . x = 111
30 . x = 111
x = \(\frac{37}{10}\)
b) 3 + ( 10 . x ) = 111
10 . x = 111 - 3
10 . x = 108
x = 108 : 10
x = 10.8
3. ( 10.x ) = 111
10.x = 111 : 3
10.x = 37
x = 37 : 10
x = 3,7
______________________
3 + ( 10.x ) = 111
10.x = 111 - 3
10.x = 108
x = 108 : 10
x = 10,8
3.(10.x)=111
=> 10x = 111 : 3
=> 10 x = 37
=> x = 37 : 10
=> x = \(\frac{37}{10}\)
3.(10.x)=111
(10.x)=111:3
(10.x)=37
x=37:10
x=3,7
vậy x=3,7
3 + (10.x) = 111
10.x = 111- 3
10.x = 108
x = 108:10
x = 10,8
3 + (10 + x ) = 111
10 + x = 111- 3
10 + x = 108
x = 108 – 10
x = 98
3.(10 + x) = 111
10+ x = 111: 3
10 + x = 37
x = 37 – 10
x = 27
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
B1. phân a tui ko bt nha :>
\(B=\frac{2^{13}\cdot9^4}{6^6\cdot8^3}\)
\(=\frac{2^{13}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\)
\(=\frac{2^{13}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
\(=\frac{2^{13}\cdot3^8}{2^{15}\cdot3^6}\)
\(=\frac{1\cdot3^2}{2^2\cdot1}\)
\(=\frac{1\cdot9}{4\cdot1}\)
\(=\frac{9}{4}\)
a)\(3.\left(10:x\right)=111\)
\(\Rightarrow10:x=37\)
\(x=10:37\)
\(x=\frac{10}{37}\)
b)\(3.\left(10+x\right)=111\)
\(\Rightarrow10+x=37\)
\(x=37-10\)
\(x=27\)
c)\(3+\left(10.x\right)=111\)
\(\Rightarrow10x=108\)
\(x=108:10\)
\(x=\frac{54}{5}\)
d)\(3+\left(10+x\right)=111\)
\(\Rightarrow10+x=108\)
\(x=108-10\)
\(x=98\)