\(\left|x+\frac{11}{67}\right|+\left|x+\frac{2}{17}\right|+\left|x+\frac{4}{17}\right|=4x\)

Có: \(\left\{{}\begin{matrix}\left|x+\frac{11}{67}\right|\\\left|x+\frac{2}{17}\right|\\\left|x+\frac{4}{17}\right|\end{matrix}\right.>0\forall x.\)

Do đó \(4x>0\Rightarrow x>0.\)

Lúc này ta có: \(\left(x+\frac{11}{67}\right)+\left(x+\frac{2}{17}\right)+\left(x+\frac{4}{17}\right)=4x\)

⇒ \(\left(x+x+x\right)+\left(\frac{11}{67}+\frac{2}{17}+\frac{4}{17}\right)=4x\)

⇒ \(3x+\frac{589}{1139}=4x\)

⇒ \(4x-3x=\frac{589}{1139}\)

⇒ \(1x=\frac{589}{1139}\)

⇒ \(x=\frac{589}{1139}:1\)

⇒ \(x=\frac{589}{1139}\left(TM\right)\)

Vậy \(x=\frac{589}{1139}.\)

Chúc bạn học tốt!

\(17x=69.17-17.67\)

\(\Leftrightarrow17x=17.\left(69-67\right)\)

\(\Leftrightarrow17x=17.2\)

\(\Leftrightarrow x=2\)

\(17x=69\times17-17\times67\)

Xét vế sau ta có: 

\(69\times17-17\times67\)

\(=\left(69-67\right)\times17\)

\(=2\times17\)

\(=34\)

Ta lại có: \(17x=34\)

\(\Rightarrow x=34:17\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

17x=69.17-17.67

17x=17.(69-67)

17x=17.2

17x=34

    x=34:17

    x=2

vậy x=2

\(67+x+y=180\) và \(y-x=17\)

=> \(x+y=180-67=113\)

Đưa về dạng tổng - hiệu

y = ( 113 + 17 ) : 2 = 65

x = 113 - 65 = 48

xD

\((x-17)+52=158\)

\(\Rightarrow x-17=158-52\)

\(\Rightarrow x-17=106\)

\(\Rightarrow x=106-17\)

\(\Rightarrow x=89\)

( x- 17 ) + 52 = 158

x - 17 = 158 - 52

x - 17 = 106

x = 106 + 7

x = 113

Vậy x = 113

( x : 23 + 45 ) : 67 = 8911

x : 23 + 45 = 8911 x 67

x : 23 + 45 = 597037

x : 23 = 597037 - 45 

x : 23 = 596992

x = 596992 x 23

x = 13730816

Vậy x = 13730876

(x-17)+52=158

x - 17 = 158 - 52

x - 17 = 106

x= 106+17

x= 123

(x :23 + 45):67=8911

(x : 68 ) : 67 = 8911

x : 68 = 8911 * 67

x : 68 = 597037

x= 597037 x 68

x = 40598516

k mình nhé !!!

45+x=67

      x=67-45

      x=22

17-x=7

     x=17-7

      x=10

x.22=44

x     =44:22

x      =2

a) x = 22

b) x = 10

c) x = 2

45+x=67

x=67—45

x=32

Lời giải:

a. $3x-2=19$

$3x=19+2=21$

$x=21:3=7$

b. $3(17-x)=6^7:6^5=6^2=36$

$17-x=36:3=12$

$x=17-12=5$

c, Ta có : x ⋮ 18 ; x ⋮ 16 và  300 < x < 400

=> x ∈ BC(18;16).

Mà 18 = 2 . 3²    ;   16 = 2⁴ 

=> BCNN(18;16) = 2⁴ . 3² = 144

=> x ∈ B(144) ∈ {0;144;288;...}

Mà 300 < x < 400 => Không có số x thoả mãn

\(a,\left(37-17\right).\left(-5\right)+23.\left(13-17\right)\)

\(=20.\left(-5\right)+23.\left(-4\right)\)

\(=-100-92\)

\(=-192\)

\(b,\left(-57\right).\left(67-34\right)-67\left(34-57\right)\)

\(=\left(-57\right).33-67.\left(-23\right)\)

\(=-1881+1541\)

\(=-340\)

a) 20.(-5)+23.(-30)

=-100+(-690)

=-790

b)(-57).33-67.(-23)

=-1881-(-1541)

=-340