-\(\dfrac{-333}{777}+\dfrac{22}{55}\)
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\(E=\dfrac{-1}{3}-\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{1}{5}=-1\)
\(\dfrac{11}{6}+\dfrac{1}{4}=\dfrac{22}{12}+\dfrac{3}{12}=\dfrac{25}{12}\)
\(\dfrac{2}{5}-\dfrac{3}{8}=\dfrac{16}{40}-\dfrac{15}{40}=\dfrac{1}{40}\)
\(\dfrac{3}{10}-\dfrac{4}{15}=\dfrac{9}{30}-\dfrac{8}{30}=\dfrac{1}{30}\)
\(3+\dfrac{2}{5}=\dfrac{15}{5}+\dfrac{2}{5}=\dfrac{17}{5}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{15}{35}+\dfrac{14}{35}=\dfrac{29}{35}\)
a) Ta có: 333777 = 333111.7 = (7773)111
777333 = 777111.3 = (7773)111
Vì 7773<3337 nên (7773)111 < (7773)111
Vậy 333777 > 777333
b) Ta có: 2222 = 22.111 =(2111)2
2222 = 2211.2 = (2211)2
Vì 2111 > 2211 nên (2111)2 > (2211)2
= 2/5 + 1/5 + 1/5
= 4/5
b. = 2/7 + 4/7 + 5/21
= 6/21 + 12/21 + 5/21
= 23/21
HT
câu a :
\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\\ \left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=\dfrac{7}{4}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}+\dfrac{1}{2}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\\ x-\dfrac{1}{3}=\sqrt{\dfrac{9}{4}}\\ x-\dfrac{1}{3}=\dfrac{3}{2}\\ x=\dfrac{3}{2}+\dfrac{1}{3}\\ x=\dfrac{11}{6}\)
câu b :
\(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\\ \Rightarrow\left(x-3\right)\cdot\left(x-3\right)=\left(-2\right)\cdot\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ x-3=\sqrt{16}\\ x-3=4\\ x=4+3\\ x=7\)
câu c :
\(\dfrac{9}{x}=\dfrac{-35}{105}\\ \Rightarrow\left(-35\right)\cdot x=9\cdot105\\ \left(-35\right)\cdot x=945\\ x=945\div\left(-35\right)\\ x=-27\)
a) \(\dfrac{22}{55}=\dfrac{2}{5}\)
b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)
c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)
d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)
=3/7+2/5=15/35+14/35=29/35
=3/7+2/5=15/35+14/35=29/35