C = ( 1/5 + 1/10 + 1/15 + .................+ 1/1000 )
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a)(-0,25).4/17.(-3 5/21).(-7/12)
= -1/17 . (-68/21) . ( -7/21)
= -4/63
b) (-2/5).4/15+(-3/10).4/15
= 4/15. [-2/5 + (-3/10)]
= 4/15 . (-7/10)
= -14/75
c)21-3 3/4:(3/8-1/6)
= 21 - 15/4 : 5/24
= 21 - 18
= 3
d) (-3/4+2/5):3/7+(3/5+-1/4):3/7
= [(-3/4 + 2/5) + (3/5 + -1/4)] : 3/7
= [-7/20 + (-17/20)] : 3/7
= -6/5 : 3/7
= -14/5
a) \(...=-\dfrac{1}{4}.\dfrac{4}{17}.\left(-\dfrac{63}{21}\right).\left(-\dfrac{7}{12}\right)\)
\(=-\dfrac{1}{17}.\dfrac{63}{21}.\dfrac{7}{12}\)
\(=-\dfrac{7}{68}\)
b) \(...=-\dfrac{2}{5}.\dfrac{4}{15}-\dfrac{3}{10}.\dfrac{4}{15}\)
\(=\dfrac{4}{15}\left(-\dfrac{2}{5}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}\left(-\dfrac{4}{10}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}.\left(-\dfrac{7}{10}\right)=-\dfrac{14}{75}\)
c) \(...=21-\dfrac{15}{4}:\left(\dfrac{9}{24}-\dfrac{4}{24}\right)\)
\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)
\(=21-\dfrac{15}{4}.\dfrac{24}{5}\)
\(=21-3.6=3\)
d) \(...=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{7}{3}\left(-1+1\right)=0\)
Câu 1: (2x-3)-(x-5)=(x+2)-(x-1)
2x -3 -x+5 = x+2 -x +1
2x -x -x +x = 2+1 +3 -5
x= 1
Câu 2: 2(x-1)-5(x+2)=10
2x -2 -5x -10 =10
2x -5x = 10 +2 +10
(2-5) x = 22
-3x= 22
x= 22/-3
Câu 1: ( 2x - 3 ) - ( x - 5 ) = ( x + 2 ) - ( x - 1 )
=> ( 2x - x ) - ( 3 - 5 ) = ( x - x ) + ( 2 + 1 )
=> x + 2 = 3
=> x = 1
Thử lại: ( 2 - 3 ) - ( 1 - 5 ) = ( 1 + 2 ) - ( 1 - 1 )
=> -1 + 4 = 3 - 0
=> 3 = 3 ( thoả mãn )
Câu 2: 2 ( x - 1 ) - 5 ( x + 2 ) = 10
=> ( 2x - 2 ) - ( 5x + 10 ) = 10
=> ( 2x - 5x ) - ( 2 + 10 ) = 10
=> -3x - 12 = 10
=> -3x = 22
=> x = -22/3
Thử lại: 2 ( -22/3 - 1 ) - 5 ( -22/3 + 2 ) = 10
=> 2 * -25/3 - 5 * -16/3 = 10
=> -50/3 - -80/3 = 10
=> (-50) - (-80)/3 = 10
=> 30 / 3 = 10 ( thoả mãn )
(2x−1)+(4x−2)+...+(400x−200)=5+10+...+1000(2x−1)+(4x−2)+...+(400x−200)=5+10+...+1000
⇒(2x+4x+...+400x)−(1+2+...+200)⇒(2x+4x+...+400x)−(1+2+...+200)
=5.(1+2+...+200)=5.(1+2+...+200)
⇒x.[(400+2).200:2]−[(200.{200+1}):2]⇒x.[(400+2).200:2]−[(200.{200+1}):2]
=5.[(200.{200+1}):2]=5.[(200.{200+1}):2]
⇒x.40200−20100=100500⇒x.40200−20100=100500
⇒40200x=120600⇒40200x=120600
⇒x=3⇒x=3
Vậy x=3.