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1,
\(75-\left(3.5^2-4.2^3\right)=75-\left(75-32\right)=75-75+32=32\)
2,
\(58.76+24.58-5.20=58\left(76+58\right)-100=58.100-100=100\left(58-1\right)=5700\)3,
\(4.5^2-3.2^3+3^3:3^2=4.5^2-3.2^3+3^{3-2}=3\left(2^3+1\right)+100=27+100=127\)4,
\(\left[5^2.6-20\left(37-2^5\right)\right]:10-10^0=\left(5^2.6-20.5\right):10-1=\left[10\left(15-10\right)\right]:10-1=5-1=4\)cái đề bn ghi ko rõ nên mk lầm thì báo mk nhé
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A, 210x13+210x65/28x104
=210x(13+65)/28x4x26
=210x78/28x22x26
=210x78/210x26
=78/26
=3
b, (1+2+3+...+100)x(12+22+32+...+102)x(65x111-13.15.37)
=(1+2+3+...+100)x(12+22+..+102)x(7215-7215)
=(1+2+..+100)x(12+22+..+102)x0
=0
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- 22.32.5:22.3-32=3.5-32=15-9=6
- 2.52-22.32:32=2.(52-2)=2.(25-2)=46
3. 33.19-33.12=33.(19-12)=33.7=189
4. 3.52-16:22=3.52-24:22=3.25-4=75-4=71
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[(37-32)^3-5^10:5^8]+2021^0
=[5^3-5^2]+1
=[125-25]+1
=100+1=101
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Câu 1.9920và 999910
=(992)10=980110
Vậy 980110<999910 suy ra 9920<999910
Câu 2. 3500và 7300
3500=(35)100=243100
7300=(73)100=343100
Vậy 243100<343100 => 3500<7300
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1. a) \(45x-37=53\)
\(45x=90\)
\(x=2\)
vay \(x=2\)
b) \(\frac{x}{9}+500=600\)
\(\frac{x}{9}=100\)
\(\frac{x}{9}=\frac{900}{9}\)
\(\Rightarrow x=900\)
vay \(x=900\)
c) \(576-x=139\)
\(x=576-139\)
\(x=437\)
vay \(x=437\)
d) \(\left(48+x\right)-27=79\)
\(48+x=79+27\)
\(48+x=106\)
\(x=58\)
vay \(x=58\)
2. \(2^5.2^7.2^9=2^{5+7+9}=2^{21}\)
\(3^9.3^2.3=3^{9+2+1}=3^{12}\)
\(10^9:10^7=10^{10-7}=10^3\)
3). \(2.2.2.2.2.2=2^6\)
\(2.3.2.2.3.3.3=2^3.3^4\)
\(10.9.10.10.10.9.9=10^4.9^3\)
1 a) x= 720
b ) x=chín trăm
c)x= 437
d)x=58
bài2 a)= 221 b)312 c ) =102
bài 3 a)=26 b)=22 nhân 34 c )= 103 nhân chín mũ 3
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\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
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\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
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1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)
2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)
\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)
3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)