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a) Ta có: \(-19-\left|+13\right|+\left|-10\right|-\left(-5\right)-11\)
\(=-19-13+10+5-11\)
\(=-28\)
b) Ta có: \(\left(-10\right)-\left(11\right)+\left|-5\right|-\left|-7\right|-8\)
\(=-10-11+5-7-8\)
\(=-31\)
c) Ta có: \(-\left(-8\right)-\left(+7\right)-\left|-11\right|+\left|+12\right|\)
\(=8-7-11+12\)
\(=2\)
d) Ta có: \(-14-\left|-18\right|-\left(-20\right)-\left|-25\right|\)
\(=-14-18+20-25\)
\(=-37\)
đ) Ta có: \(23-\left|-15\right|-\left(-17\right)+\left|-13\right|\)
\(=23-15+17+13\)
\(=38\)
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2:
=>27:3^x=2*25-16-31=50-47=3
=>3^x=27/3=9
=>3^x=3^2
=>x=2
1:
b: \(=16\cdot55+16\cdot45-1\)
=16(55+45)-1
=1600-1
=1599
c: \(=\dfrac{1800}{49-\left[2\cdot\left(36-34\right)^3-5\right]}\)
\(=\dfrac{1800}{49-2\cdot2^3+5}=\dfrac{1800}{49-16+5}=\dfrac{1800}{38}\)=900/19
d: \(=\dfrac{5\cdot3^{11}\cdot2^{11}-3^{11}\cdot2^{11}}{2^{10}\cdot3^{10}\cdot2^2\cdot3+7\cdot2^{12}\cdot3^{12}}\)
\(=\dfrac{3^{11}\cdot2^{11}\left(5-1\right)}{2^{12}\cdot3^{11}\left(1+7\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{4}{1+21}=\dfrac{4}{22\cdot2}=\dfrac{1}{11}\)
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a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)
\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)
\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)
a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)
b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)
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a) 35
b) -53
c) 40 – - 7 2 = 40 – 49 = -9
d)| -15 + 21| - | 4 – 11 | = | 6 | - | -7 | = 6 – 7 = -1
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49+(11-25).(-4)=49+(-14).(-4)
=49+56=105
\(49+\left(11-25\right)\left(-4\right)\)
\(=49-14\left(-4\right)\)
\(=49+56\)
\(=105\)
a) 35