=> x:3,24-5,87=1,41

=> x:3,24=1,41+5,87

=> x=7,28x3,24

=> x=23,5872

\(1,405=\frac{1405}{100}=1405\%\)

Bài 2

x : 3,24 - 5,87 = 2,82 x 0,5

x : 3,24 - 5,87 = 14,1

x : 3,24 = 14,1 + 5,87

x : 3,24 = 19,97

x = 19,97 x 3,24

x = 64,7028

x : 3,24 - 5,87 = 2,82 x 0,5

x : 3,24          = 1,41 

x : 3,24          =1,41 + 5,87

x                   =7,28

x                   = 7,28 x 3,24

x                   =23,5872 

Ý 1 sai đề bài là cái chắc 

Ý 2:x:3,24-5,87=2,82×0,5

x:3,24 -5,87=1,41

x:3,24=1,41+5,87

x:3,24=7,28

    x=7,28:3,24

     x=182/81

k  cho mik ikkkk

2 - ( 10,53 : 4,5 - 2,33 )  81, 9 =2 - 0,01 nhân 81,9

                                                 =1,99 nhân 81,9

                                                 = 162,981

( lưu ý dấu nhận mình không biết ghi nên là nhân )

x = 23,5872 . 

95% . 

x:3,24-5,87=2,82x0,5

x:3,24-5,87=1,41

x:3,24=1,41+5,87

x:3,24=7,28

x=7,28x3,24

x=23,5872

bạn diễn giải cho mình bài tính đi

x:3,24 - 5,87 = 2,82 . 0,8

x = [ ( 2,82.0,8 ) +5,87 ] . 3,24

x  = 26,32824

x:3,24-5,87=2,82x0,8

x:3,24-5,87=2,256

x:3,24       = 2,256 + 5,87

x:3,24       = 8,126

x              = 8,126 * 3,24

x              = 26,32824

tích mình nha !!!

x x 10 -x = 1

x x 9 = 1

x =1:9

x =1/9

1/9 là một phần chín nhé

thank you nha!!!

= 20 vì \(\frac{4}{20}=\frac{4:4}{20:4}=\frac{1}{5}\)

20 nhe

Vi 4:4=1 nên x giảm 5 lần.suy ra x=20

Sửa : \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2002}\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

Vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\ne0\)

\(\Rightarrow x+2004=0\)

=> x = -2004 

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{3}{2009}\)

\(\Rightarrow\frac{3}{3\left(x+1\right)}=\frac{3}{2009}\)

\(\Rightarrow3\left(x+1\right)=2009\)

\(\Rightarrow3x+3=2009\)

\(\Rightarrow3x=2006\)

\(\Rightarrow x=\frac{2006}{3}\)