Bài 1:Thực hiện một cách hợp lí
a) 3/4 X 5/4 - 3/5 X 1/4
b)6,3 + (-6,3) + 4,9
c) 42 x 5,43 + 5,43 x (-5)2 +59 x 5,43
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Thay giá trị
\(\frac{\left(7\sqrt{5201}\right)^4}{1600000000}\)
a ) 1,2 : x + 2,3 : x = 4
( 1,2 + 2,3 ) : x = 4
3,5 : x = 4
x = 3,5 : 4
x = 0,875
b ) 9 : x - 5,43 : x = 3
( 9 - 5,43 ) : x = 3
3,57 : x = 3
x = 3,57 : 3
x = 1,19
`#3107.101107`
`a)`
`4,57 \times 42 + 5,43 \times 42`
`= 42 \times (4,57 + 5,43)`
`= 42 \times 10`
`= 420`
`b)`
`8,4 \times 96 + 1,6 \times 96`
`= 96 \times (8,4 + 1,6)`
`= 96 \times 10`
`= 960`
a) 4,57x42+5,43x42
= (4,57+5,43)x42
= 10x42
=420
b) 8,4x96+1,6x96
= (8,4+1,6)x96
= 10x96
= 960
Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
54,3 x 5,4 + 5,43 : 0,1 x 0,3 + 54,3 x \(\dfrac{43}{10}\)
= 54,3 x 5,4 + 5,43 x 10 x 0,3 + 54,3 x 4,3
= 54,3 x 5,4 + 54,3 x 0,3 + 54,3 x 4,3
= 54,3 x (5,4 + 0,3 + 4,3)
= 54,3 x 10
= 543
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
a, 27.5 x 5.43 + 4.57 x 27.5
= 27.5 x ( 5.43 + 4.57 )
= 27.5 x 10
= 275
c,\(\frac{2006x2005-1}{2004x2006+2005}\)
= \(\frac{2006x\left(2004+1\right)-1}{2004x2006+2005}\)
= \(\frac{2006x2004+2006-1}{2004x2006+2005}\)
= \(\frac{2006x2004+2005}{2004x2006+2005}\)
= 1.
Mình chỉ biết từng nấy thôi.
*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)