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b) \(\frac{4}{10}+\frac{-2}{9}+\frac{-3}{-5}+\frac{21}{-27}+\frac{-10}{20}\)
\(=\frac{2}{5}-\frac{2}{9}+\frac{3}{5}-\frac{7}{9}-\frac{1}{2}\)
\(=\left(\frac{2}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}+\frac{7}{9}\right)-\frac{1}{2}\)
\(=1-1-\frac{1}{2}=-\frac{1}{2}\)
a) \(-\frac{5}{2}+\frac{1}{7}+\frac{6}{7}+\frac{1}{2}+\frac{3}{4}\)
\(=\left(\frac{-5}{2}+\frac{1}{2}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\frac{3}{4}\)
\(=-2+1+\frac{3}{4}\)
\(=-1+\frac{3}{4}\)
\(=-\frac{1}{4}\)
13/50+9/100+41/100+12/50
=(13/50+12/50)+(9/100+41/100)
=1/2+1/2
=1
11) Ta có:
\(\frac{120-0,5.40.5.0,2.20.0,25-20}{1+5+9+...+33+37}\)
\(=\frac{120-\left(0,5.40\right).\left(5.0,2\right).\left(20.0,25\right)-20}{1+5+9+...+33+37}\)
\(=\frac{120-20.1.5-20}{1+5+9+...+33+37}\)
\(=\frac{120-100-20}{1+5+9+...+33+37}\)
\(=\frac{0}{1+5+9+...+33+37}=0\)
a) 1152-(374+1152)+(-65+374)
=1152-374-1152-65+374
=(1152-1152)+(-374+374)-65
=-65
a, 1152 - ( 374 + 1152 ) + ( -65 + 374 )
= 1152 - 374 - 1152 - 65 + 374
= ( 1152 - 1152 ) - ( 374 - 374 ) - 65
= 0 - 0 - 65
= -65
b, 13 - 12 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 + 2 - 1
= ( 13 - 12 ) + 11 + ( 10 - 9 ) + ( 8 - 7 ) - ( 6 - 5 ) - ( 4 - 3 ) + ( 2 - 1 )
= 1 + 11 + 1 + 1 - 1 - 1 + 1
= 13
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
Bài 2 :
Ta có:
A) | 2 + 3x | = | 4x - 3 |
<=> 2 + 3x = 4x - 3
<=> 3x - 4x = -3 - 2
=> -x = -5
=> x= 5
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
a ) − 10 3 + 13 10 − 1 6 + 7 10 = 13 10 + 7 10 + − 10 3 + − 1 6 = 2 − 7 2 = − 3 2
b ) 1 − 20 6 + 39 30 − 4 24 + 4 24 = 1 − 10 3 + 13 10 − 1 6 + 7 10 = − 1 2