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a) 5.8+7=47 ; 5.(8+7)=75
Nên 47 < 75.
b) 7.15-5=100 ; 7.(15-5)=70
Nên 100 > 70.
a, 5.7+7 và 5.(8+7)
=> 5.7+7 và 5.8+35
=> ...
b, 7.15-5 và 7.(15-5)
=> 7.15-5 và 7.15- 35
=> ...
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-7}{21}+1+\frac{1}{3}\)
\(A=\left(\frac{-7}{21}+\frac{1}{3}\right)+1\)
\(A=0+1\\ A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)
\(B=\frac{2}{15}+\frac{-1}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\left(\frac{3}{12}+\frac{-3}{4}\right)\)
\(C=\frac{-1}{5}+\frac{-1}{2}\)
\(C=\frac{-7}{10}\)
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-1}{3}+1+\frac{1}{3}\)
\(A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-6}{9}\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-10}{15}\)
\(B=\frac{-8}{15}+\frac{5}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\frac{3}{4}+\frac{-3}{4}\)
\(C=\frac{-1}{5}\)
a) c1 := -2/-5+-5/-6+4/5
=2/5+5/6+4/5
=12/30+25/30+24/30
=61/30
c2:=(2/5+4/5)+4/5
=(2/5+4/5)+5/6
=6/5+5/6
=36/30+25/30
=61/30
b)c1:=-3/-4+11/-15+-1/2
=3/4+-11/15+-1/2
=45/60+-44/30+-30/60
-29/30
c2:=3/4+(-11/15+-1/2)
=(3/4+-1/2)+-11/15
=(3/4+-2/4)+-11/5
=1/4+-11/15
=15/60+-44/60
=-29/60
a) Thế x và y ta có:
\(-2.\left(-3\right)-5+11+3.\left(-3\right)\)
\(=6-5+11-9=3\)
b) Thế x và y ta có:
\(2.5-3.\left(-3\right)+5\left(5-\left(-3\right)\right)+15\)
\(=10+9+5\left(5+3\right)+15\)
\(=10+9+40+15=74\)
c) Thế x và y ta có:
\(4.\left(-3\right)-4\left(-3-2.5\right)-7\left(5-2\right)\)
\(=-12-4.\left(-13\right)-7.3\)
\(=-12+52-21=19\)
a) Cách 1:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \frac{2}{5} + \frac{5}{6} + \frac{4}{5}\\ = \frac{{12}}{{30}} + \frac{{25}}{{30}} + \frac{{24}}{{30}} = \frac{{61}}{{30}}\end{array}\)
Cách 2:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \left( {\frac{2}{5} + \frac{4}{5}} \right) + \frac{5}{6}\\ = \frac{6}{5} + \frac{5}{6} = \frac{{36}}{{30}} + \frac{{25}}{{30}} = \frac{{61}}{{30}}\end{array}\)
b) Cách 1:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \frac{{45}}{{60}} + \frac{{ - 44}}{{60}} + \frac{{ - 30}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\).
Cách 2:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \left( {\frac{3}{4} + \frac{{ - 1}}{2}} \right) + \frac{{ - 11}}{{15}}\\ = \left( {\frac{3}{4} + \frac{{ - 2}}{4}} \right) + \frac{{ - 11}}{{15}}\\ = \frac{1}{4} + \frac{{ - 11}}{{15}}\\ = \frac{{15}}{{60}} + \frac{{ - 44}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\)
tính giá trị biểu thức
C=1/15+1/35+.................+1/2499
B=5/1.4+5/4.7+...................5/100.103
b: \(7\cdot2^{13}< 8\cdot2^{13}=2^{16}\)
d: \(3^{99}=\left(3^{33}\right)^3\)
\(11^{21}=\left(11^7\right)^3\)
mà \(3^{33}>11^7\)
nên \(3^{99}>11^{21}\)