(8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2

<=>144 x y -90 x y -54 x y + 2 x y = 58

<=>2 x y = 58

<=> y = 29

Vậy y = 29

Tìm y biết : 
(8 x 18 - 5 x 18 - 18 x 3) x y + 2 x y = 8 x 7 + 2

                    [ 18 x ( 8 - 5 - 3 ) ] x y = 56 + 2

                    [ 18 x ( 8 - 5 - 3 ) ] x y = 58

                                 ( 18 x 0 ) x y = 58

                                 0 x y + 2 x y = 58

                                            2 x y = 58 

                                                  y = 58 : 2

                                                  y = 29

(8 x 18 - 5 x 18 - 18 x 3) x y + 2 x y

=[(8 - 5- 3) x 18] x y + 2 x y

=[0 x 18] x y + 2 x y

= 0 nhân cho bất kì số nào cũng bằng 0 nên chứng tỏ bài này cho sai đề 

(8 x 18 - 5 x 18 - 18 x 3)x + 2x = 8 x 7 + 24

<=> [18 x (8 - 5 - 3)]x + 2x =   80

<=> 2x = 80

<=> x = 40

      (8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2

<=>144 x y -90 x y -54 x y + 2 x y = 58

<=>2 x y = 58

<=> y = 29

vậy y = 29

( mình ko chép lại đề bài đâu nha ,giải lun đó)

[18x(8-5-3)]xy+2xy=56+2

(18x0)xy+2xy=58

0xy+2xy=58

2xy=58

  y=58:2=29

tick cho mình nha

\(2 .x + 4 .x + . . . + 18 .x = 18 + 16 + . . . + 2 \)

x=1

x=1

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

a) 32 : 4 : 2 = 8 : 2 = 4 

     32 : 4 : 2 = 32 : 2 = 16 

b) 18 : 2 x 3 = 18 : 6 = 3

     18 : 2 x 3 = 9 x 3 = 18

lớp 3???

help

help

a, \(\left|x-1\right|=20\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=20\\x-1=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=-19\end{matrix}\right.\)

b, đk : x =< 18/3 

\(\Leftrightarrow\left[{}\begin{matrix}x-2=18-3x\\x-2=3x-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=8\left(ktm\right)\end{matrix}\right.\)

c, <=> | x - 2 | = 18 - 4x 

đk : x =< 18/4 = 9 /2 

\(\Leftrightarrow\left[{}\begin{matrix}x-2=18-4x\\x-2=4x-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{16}{3}\left(ktm\right)\end{matrix}\right.\)

\(\left|x-1\right|-8=12\)

\(\left|x-1\right|=12+8\)

\(\left|x-1\right|=20\)

\(x-1=+-20\)

a)\(\left|x-1\right|-8=12\Rightarrow\left|x-1\right|=20\)

   \(\Rightarrow\left[{}\begin{matrix}x-1=20\\x-1=-20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=21\\x=-19\end{matrix}\right.\)

b)\(\left|x-2\right|=18-3x\)

   \(\Rightarrow\left[{}\begin{matrix}x-2=18-3x\\x-2=3x-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=20\\-2x=-16\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=8\end{matrix}\right.\)

c)\(\left|x-2\right|-18+4x=0\)

   \(\Leftrightarrow\left|x-2\right|=18-4x\)

   Làm tương tự câu b

\(x\sqrt{18}-\sqrt{18}=x\sqrt{18}+4\sqrt{2}\\ \Rightarrow-3\sqrt{2}=4\sqrt{2}\left(vô.lí\right)\)

\(x\sqrt{18}-\sqrt{18}=x\sqrt{18}+4\sqrt{2}.\)

\(\Leftrightarrow x\sqrt{18}-\sqrt{18}-x\sqrt{18}=4\sqrt{2}.\)

\(\Leftrightarrow-3\sqrt{2}=4\sqrt{2}\) (vô lý).

<=> x\(\sqrt{18}\) - x\(\sqrt{18}\) = \(4\sqrt{2}\) + \(\sqrt{18}\)

<=> 0x = \(4\sqrt{2}\) + \(3\sqrt{2}\)

<=>0x = \(7\sqrt{2}\)