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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
B1:
\(\dfrac{3}{4}=\dfrac{3\times10}{4\times10}=\dfrac{30}{40}=\dfrac{75}{100};\dfrac{4}{5}=\dfrac{4\times8}{5\times8}=\dfrac{32}{40}=\dfrac{80}{100}\\ Vì:\dfrac{30}{40}< \dfrac{31}{40}< \dfrac{32}{40}.Nên:\dfrac{3}{4}< \dfrac{31}{40}< \dfrac{4}{5}\\ Và:\dfrac{75}{100}< \dfrac{77}{100}< \dfrac{79}{100}< \dfrac{80}{100}.Nên:\dfrac{3}{4}< \dfrac{77}{100}< \dfrac{79}{100}< \dfrac{4}{5}\)
3 phân số nằm giữa 2 phân số \(\dfrac{3}{4}\) và \(\dfrac{4}{5}\) là: \(\dfrac{31}{40};\dfrac{77}{100};\dfrac{79}{100}\)
B2:
\(\dfrac{3}{5}=\dfrac{3\times2}{5\times2}=\dfrac{6}{10};\dfrac{4}{5}=\dfrac{4\times2}{5\times2}=\dfrac{8}{10}\)
Vì: 6<7<8. Nên phân số có mẫu số bằng 10, lớn hơn \(\dfrac{3}{5}\) và nhỏ hơn \(\dfrac{4}{5}\) là \(\dfrac{7}{10}\)
Bài 1:
Tổng của 2 số là
\(36\times2=72\)
Số lớn là
\(72-17=55\)
Bài 2:
a) \(4567+y\div34=10987\)
\(y\div34=10987-4567\)
\(y\div34=6420\)
\(y=6420\times34\)
\(y=218280\)
b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\)
\(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\)
\(\dfrac{1}{2}\div y=\dfrac{2}{3}\)
\(y=\dfrac{1}{2}\div\dfrac{2}{3}\)
\(y=\dfrac{3}{4}\)
Bài 3:
a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\)
b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)
Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :
(a+b):2=36⇒a+b=72
mà b=17
Nên a=72-17=55
Bài 2 :
a) 4567+y:34=10987
⇒ y:34=10987-4567
⇒ y:34=6420
⇒ y=6420x34
⇒ y=218280
b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)
\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)
\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)
\(\Rightarrow y=\dfrac{3}{4}\)
Bài 3 :
\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)
= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)
\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)