(x-45).27 = 0

=> x - 45 = 0

=> x = 45 + 0

=> x = 45

23.(42 - x) = 23

=> 42 - x = 23 : 23

=> 42 - x = 1

=> x = 42 - 1

=> x = 41

l--i-k-e cho mk

a/ => x - 45 = 0 => x = 45

b/ => 42 - x = 1 => x = 41

a) x=45

b) x=41

a) ( x -45 ) .27 =0                                                 b)  23 . (42 -x ) = 23

    ( x -45 ) = 0                                                                  42 - x = 1                                                                

           x   = 45                                                                        x = 41

Nhớ kb và tk cho mk nhé . 

\(\text{a , x = 25 }\)

\(\text{b, x = 41}\)

a) (x - 45) .27 = 0 

=> x - 45 = 0

=> x = 45

b) 23 . (42 - x) = 23

42 - x = 1

x = 42 - 1

x = 41

a) x=45

b) x=41

k mình nha

a) x= 45

b) x= 41

a) ( x - 45 ) . 27 = 0

( x - 45 )        = 0 : 27

( x - 45 )        = 0

  x                 = 0 + 45

  x                 = 45

b) 23 . ( 42 - x ) = 23

          ( 42 - x ) = 23 : 23

          ( 42 - x ) = 1

                   x = 42 - 1

                    x = 41

\(\text{Ta có: (x - 45).27 = 0 }\) 
\(\text{ x - 45 = 0 : 27 }\) 
\(\text{ x - 45 = 0 }\)         
\(\text{ x = 0 + 45 }\)       

\(\text{ x = 45}\)

a) (x-45).27=0

=>x-45=0

=>x=0+45

=>x=45

b)23.(42-x)=23

=>42-x=23:23

=>42-x=1

=>x=42-1

=>x=41

a) (x - 45) . 27 = 0

=> x - 45 = 0

x = 45

Vậy x = 45.

b) 23 (42 - x) = 23

42 - x = 23 : 23

42 - x = 1

x = 42 - 1

x = 41

Vậy x = 41.

a) (x – 45).27 = 0

=>   x -  45  =  0

=>  x  =  45

 b) 23.(42- x) = 23 

=>  42- x  =  1

=>   x  =  41

c. 3x – 5=7

=>  3x  =  12

=>  x  =  4

e. 15 – 5x=10

=>  5x  =  5

=>  x  =  1

Ta có:

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)

\(\frac{11}{45}.x=\frac{23}{45}\)

 \(x=\frac{23}{45}\div\frac{11}{45}\)

\(x=\frac{23}{11}\)

=> \(x=\frac{23}{11}\)

\(23-y^2=7\left(x-2004\right)^2\ge0\\ \Leftrightarrow y^2\le23\)

Mà \(y\in N\Leftrightarrow y\in\left\{0;1;2;3;4\right\}\)

Với \(y=0\Leftrightarrow7\left(x-2004\right)^2=23\left(loại\right)\)

Với \(y=1\Leftrightarrow7\left(x-2004\right)^2=22\Leftrightarrow\left(x-2004\right)^2=\dfrac{22}{7}\left(loại\right)\)

Với \(y=2\Leftrightarrow7\left(x-2004\right)^2=19\Leftrightarrow\left(x-2004\right)^2=\dfrac{19}{7}\left(loại\right)\)

Với \(y=3\Leftrightarrow7\left(x-2004\right)^2=14\Leftrightarrow\left(x-2004\right)^2=2\left(loại\right)\)

Với \(y=4\Leftrightarrow7\left(x-2004\right)^2=7\Leftrightarrow\left[{}\begin{matrix}x-2004=1\\x-2004=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2005\\x=2003\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2005;4\right);\left(2003;4\right)\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{46}{45}\)

\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x\)

\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right).x\)

\(=\left(\frac{1}{2}-\frac{1}{90}\right).x=\left(\frac{45}{90}-\frac{1}{90}\right)x=\frac{44}{90}.x=\frac{22x}{45}=\frac{46}{45}\)

=> 22x=46

=> x=\(46:22=\frac{23}{11}\)