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1) 40 + 15 + (-10) + (-15) 2) -13 + (-750) + (-17) + 750 3) (35 - 17) + (17 + 120 - 35)
= 40 + 15 - 10 - 15 = -13 - 750 - 17 + 750 = 35 - 17 + 17 + 120 - 35
= (40 - 10) + (15 - 15) = (-13 - 17) + (-750 + 750) = (35 - 35) + (-17 + 17) + 120
= 30 = -30 = 120
4) (55 + 45 + 15) - (15 - 55 + 45) 5) -(12 + 21 - 23) - (23 - 21 + 10) 6) (2020 - 79 + 15) - (-79 + 15)
= 55 + 45 + 15 - 15 + 55 - 45 = -12 -21 + 23 - 23 + 21 - 10 = 2020 - 79 + 15 + 79 - 15
= (45 - 45) + (15 - 15) + (55 + 55) = (-12 - 10) + (-21 + 21) + (23 - 23) = 2020 + (-79 + 79) + (15 - 15)
= 110 = -22 = 2020
7) -(515 - 80 + 91) - (2010 + 80 - 91) 8) 25 - (-17) + 24 - 12 9) 235 - (34 + 135) - 100
= -515 + 80 - 91 - 2010 - 80 + 91 = 25 + 17 + 24 - 12 = 235 - 34 - 135 - 100
= (-515 -2010) + (80 - 80) + (-91 + 91) = 54 = -34
= -2525
10) (13 + 49) - (13 - 135 + 49)
= 13 + 49 - 13 + 135 - 49
= (13 - 13) + (49 - 49) +135
= 135
a) \(15.4=15.\left(2.2\right)=\left(15.2\right).2=30.2=60\)
\(25.12=25.\left(2.6\right)=\left(25.2\right).6=50.6=300\)
\(125.16=125.\left(2.8\right)=\left(125.8\right).2=1000.2=2000\)
b) \(25.12=25.\left(10+2\right)=25.10+25.2=250+50=300\)
\(34.11=34.\left(10+1\right)=34.10+34.1=340+34=374\)
\(47.101=47.\left(100+1\right)=47.100+47.1=4700+47=4747\)
Chúc em học tốt!!!
B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)
C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
45/7 : 3/5 + 1 = 82/7
3/5 x 8/9 = 8/15
\(-\dfrac{1}{45}-\dfrac{3}{5}=-\dfrac{1}{45}-\dfrac{27}{45}=\dfrac{-1-27}{45}=\dfrac{-28}{45}\)