ai biết làm bài này giúp mình với ạ
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\(\left(\dfrac{2}{7}-\dfrac{9}{4}\right)-\left(-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{9}{4}+\dfrac{3}{7}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\left(\dfrac{9}{4}+\dfrac{5}{4}+\dfrac{2}{4}\right)\)
\(=2-4\)
\(=-2\)
\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(\dfrac{5}{3}-\dfrac{3}{2}\right)+\left(\dfrac{7}{3}-\dfrac{5}{2}\right)\\ =\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}\\ =\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)+\left(-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)\\ =-\dfrac{1}{2}+\dfrac{1}{3}\\ =-\dfrac{1}{6}\)
\(a)4^7:2^5\\ =\left(2^2\right)^7:2^5\\ =2^{14}:2^5\\ =2^9\\ b)3^{10}:9^3\\ =3^{10}:\left(3^2\right)^3\\ =3^{10}:3^6\\ =3^4\\ c)27^9:3^{10}\\ \left(3^3\right)^9:3^{10}\\ =3^{27}:3^{10}\\ =3^{17}\\ d)25^5:5^3\\ =\left(5^2\right)^5:5^3\\ =5^{10}:5^3\\ =5^7\\ e)36^7:6^4\\ =\left(6^2\right)^7:6^4\\ =6^{14}:6^4\\ =6^{10}\\ g)4^3\cdot8^4\\ =\left(2^2\right)^3\cdot\left(2^3\right)^4\\ =2^6\cdot3^{12}\\ =2^{18}\)
Ta có :
\(12=2^2.3\)
\(15=3.5\)
\(=>BCNN\left(12;15\right)=3.5.2^2=3.5.4=60\)
\(=>60:12=5;60:15=4\)
\(\dfrac{5}{12}=\dfrac{5.5}{12.5}=\dfrac{25}{60}\)
\(\dfrac{8}{15}=\dfrac{8.4}{15.4}=\dfrac{32}{60}\)
Vì \(25< 32\) nên
\(=>\dfrac{25}{60}< \dfrac{32}{60}\)
\(=>\dfrac{5}{12}< \dfrac{8}{15}\)
Vậy \(\dfrac{5}{12}< \dfrac{8}{15}\)
Nếu có gì sai sót thì nhớ bảo mình , mình cảm ơn!
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Nguyên tử chlorine gồm hạt nhân có 17 proton, 18 neutron và lớp vỏ gồm 17 electron.
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Khối lượng của nguyên tử chlorine là 35.45 amu.
2B:
a) C1: \(\dfrac{-7}{12}=\dfrac{-6-1}{12}=\dfrac{-6}{12}+\dfrac{-1}{12}=\dfrac{-1}{2}+\dfrac{-1}{12}\)
C2: \(\dfrac{-7}{12}=\dfrac{-3-4}{12}=\dfrac{-3}{12}+\dfrac{-4}{12}=\dfrac{-1}{4}+\dfrac{-1}{3}\)
C4: \(\dfrac{-7}{12}=\dfrac{-2-5}{12}=\dfrac{-2}{12}+\dfrac{-5}{12}\)
b) C1: \(\dfrac{-7}{12}=\dfrac{4-11}{12}=\dfrac{4}{12}-\dfrac{11}{12}=\dfrac{1}{3}-\dfrac{11}{12}\)
C2: \(\dfrac{-7}{12}=\dfrac{2-9}{12}=\dfrac{2}{12}-\dfrac{9}{12}=\dfrac{1}{6}-\dfrac{3}{4}\)
C3: \(\dfrac{-7}{12}=\dfrac{3-10}{12}=\dfrac{3}{12}-\dfrac{10}{12}=\dfrac{1}{4}-\dfrac{5}{6}\)
Bài 1B:
a)
\(\dfrac{-1}{16}+\dfrac{-1}{24}\\ =\dfrac{-3}{48}+\dfrac{-2}{48}\\ =\dfrac{-5}{48}\)
b)
\(\dfrac{-1}{8}-\dfrac{3}{20}\\ =\dfrac{-5}{40}-\dfrac{6}{40}\\ =\dfrac{-11}{40}\)
c)
\(-\dfrac{18}{10}+0,4\\ =\dfrac{-9}{5}+\dfrac{2}{5}\\ =\dfrac{-7}{5}\)
d)
\(6,5-\left(-\dfrac{1}{5}\right)\\ =\dfrac{13}{2}+\dfrac{1}{5}\\ =\dfrac{65}{10}+\dfrac{2}{10}\\ =\dfrac{67}{10}\)
\(\dfrac{x-2y}{z-y}=-5\Rightarrow\dfrac{x-2y}{y-z}=5\\ \Rightarrow x-2y=5\left(y-z\right)\\ \Rightarrow x-2y=5y-5z\\ \Rightarrow x+5z=7y\)
Ta có:
\(\dfrac{1}{7}\cdot\dfrac{x-2z}{y-z}=\dfrac{x-2z}{7\left(y-z\right)}=\dfrac{x-2z}{7y-7z}\\ =\dfrac{x-2z}{x+5z-7z}=\dfrac{x-2z}{x-2z}=1\)
\(\Rightarrow\dfrac{x-2z}{y-z}=1:\dfrac{1}{7}=7\)