a) \(x+215=480\)

\(x=480-215\)

\(x=265\)

b) \(725-x=185:5\)

\(725-x=37\)

\(x=725-37\)

\(x=688\)

c) \(x\times\dfrac{1}{3}+x\times\dfrac{4}{3}=\dfrac{2}{3}\)

\(x\times\left(\dfrac{1}{3}+\dfrac{4}{3}\right)=\dfrac{2}{3}\)

\(x\times\dfrac{5}{3}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\times\dfrac{3}{5}\)

\(x=\dfrac{2}{5}\)

d) \(\left(\dfrac{13}{5}+\dfrac{2}{3}\right):x=5\)

\(\dfrac{49}{15}:x=5\)

\(x=\dfrac{49}{15}:5=\dfrac{49}{15}\times\dfrac{1}{5}\)

\(x=\dfrac{49}{75}\)

a)x + 215 = 480 

x = 480 - 215 

x = 265 

b) 725 - x = 185 : 5 

    725 - x = 37

              x = 725 - 37 

             x = 688

c, \(x\) \(\times\) \(\dfrac{1}{3}\) + \(x\) \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{2}{3}\)

    \(x\) \(\times\)( \(\dfrac{1}{3}\) + \(\dfrac{4}{3}\)) = \(\dfrac{2}{3}\)

    \(x\) \(\times\) \(\dfrac{5}{3}\)          = \(\dfrac{2}{3}\)

    \(x\)                  = \(\dfrac{2}{3}\) : \(\dfrac{5}{3}\)

    \(x\)                  = \(\dfrac{2}{5}\)

d, (\(\dfrac{13}{5}\) + \(\dfrac{2}{3}\)) : \(x\) = 5

     \(\dfrac{49}{15}\) : \(x\)        = 5

             \(x\)           = \(\dfrac{49}{15}\) : 5

              \(x\)         = \(\dfrac{49}{75}\)

Theo mình kết quả là

a, X là 4     b, X = 4    c, X = 5    d, X = 3    e, X = 16

Đúng thì k nhé . 

THANKS

a) x=4

b) x=4

c) x=4

d) x=-11

e) x=26

a](x-4)=25.0=x-4=0       x=0+4=4

a) ( x - 4 ) . ( x - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc x - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}\)

b) ( x - 4 ) . ( y - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc y - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4,x=3\\y=4,y=3\end{cases}}\)

c) x - 24 : 12 = 15

x - 2 = 15

x = 13

d) ( x - 12 ) : 13 = 5

x - 12 = 65

x = 77

e) 1 + 3 + 5 + ..... + x = 36

..... + x = 27

..... + x = 20

x = 11

f) 2 + 4 + .... + 2x = 110

2 + 4 + ....x = 55

Thỏa mãn điều kiện \(x\in N\left|2x+0\right|x⋮2\)

\(x\in\varnothing\)

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

thanks các bn

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

Tìm x biết:

a,x-1/2 =2/15

x=2/15 + 1/2

x =19/30

Vậy x =19/30

b,5 và 3/4 +x =15 - 2 và 1/2

23/4 + x  =27/2

x =27/2 - 23/4

x=31/4

Vậy x =31/4

c,x : 14,1 =0,25

x =0,25.14,1

x=3,525

Vậy x = 3.525

d,x + 1/2.1/3 =3/4

x + 1/6 =3/4

x =3/4 - 1/6

x =7/12

Vậy x = 7/12

a)\(x=\frac{2}{15}+\frac{1}{2}=\frac{4+1}{15x2}=\frac{5}{30}=\frac{1}{6}\)

b)anh chưa hiểu lắm hỗn số à em

c)\(x=\frac{1}{4}x\frac{141}{10}=\frac{141}{40}=3,525\)

d)\(x+\frac{1}{6}=\frac{3}{4}\)

suy ra: \(x=\frac{3}{4}-\frac{1}{6}=\frac{18-4}{24}=\frac{14}{24}=\frac{7}{12}\)

a x - 1\2 = 2\15

x            =2\15 + 1\2

x            =19\30

b 23\4 + x =27\2

x               =27\2-23\4

x                =31\4

c x :14,1=0,25

x            =0,25 x 14 ,1

x            =3,525

d x +1\2 .1\3 =3\4

x+1\6 =3\4

x        =3\4-1\6

x        =7\12

a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)

=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)

=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)

=> \(x-\frac{11}{5}=2\)

=> \(x=2+\frac{11}{5}\)

=> \(x=\frac{21}{5}\)

thanks bn

1/15 +1/35 + 1/63 + ... + 1/3135

= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57

= 1/3 - 1/57

= 18/57

1/3 + 1/6 + 1/10 +  ... + 1/x(x + 1)*2 = 2009/2011

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011

=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1)  = 2009/2011

=> 2(1/2 - 1/x + 1) = 2009/2011

=> 1/2 - 1/x + 1 = 2009/4022

=> 1/x + 1 = 1/2011

=> x + 1 = 2011

=> x = 2010

vậy-

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=1-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)