Chọn C

(x-342):6=0

(x-342)=0x6

x-342=0

x=0+342

x=342

k mk nhaVuTHi Huyen

Ta có : (x - 342) : 6 = 0

=> x - 342 = 6

=> x = 6 + 342

=> x = 348

 (x - 342) : 6 = 0

=> x - 342 = 6

=> x = 6 + 342

=> x = 348

đâu ^ sẽ là số 3

thì có thể thỏa mãn bài đề ra

k nha cảm ơn

4^x+342=7^y

4^x  phải lẻ vì 7^y lúc nào cũng lẻ

=> x =0 ( 4^0 = 1 ; 1 lẻ )

có 7^y=342+1

=> 7^y = 343

=> 7^3=343

=> y =3

K nhé

           478:9=53,11

           230:5=46

        2. x=87,7

            x=3,32

            x=284

           k mk vs ;)

            Hok tốt !!!

1

342:5=342/5        478:9=478/9          230:5=46

2

X x9=789       X=789:9       x=263/3

186:X=56    X=186:56      X=93/28

675+X=959     X=959-675    X=284

1.

342:5=68,4

478:9=53,53,1111

230:5=46

2

a,x*9=789

x=789:9

x=87,6666.

b,186:x=56

x=186:56

x=3.3214.

c,675+x=959

x=959-675

x=284

hok tốt

k mik nha

\(x\left(x+1\right)=156\)

\(\Rightarrow x^2+x=156\)

\(\Rightarrow x^2+x-156=0\)

\(\Rightarrow x^2+13x-12x-156=0\)

\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)

\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)

___________________

\(x\left(x+1\right)=342\)

\(\Rightarrow x^2+x=342\)

\(\Rightarrow x^2+x-342=0\)

\(\Rightarrow x^2+19x-18x-342=0\)

\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)

\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)

__________________

\(x\left(x+1\right)=650\)

\(\Rightarrow x^2+x=650\)

\(\Rightarrow x^2-x+650=0\)

\(\Rightarrow x^2+26x-25x-650=0\)

\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)

\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)

______________________

\(x\left(x+1\right)=380\)

\(\Rightarrow x^2+x=380\)

\(\Rightarrow x^2+x-380=0\)

\(\Rightarrow x^2+20x-19x-380=0\)

\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)

\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)

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

a, \(x\).(\(x\) + 1) = 156

156 = 2².3.13 = 12.13

Vậy \(x\).(\(x\) + 1) = 12.13

Vậy \(x\) = 12

b, \(x.\)(\(x\) + 1) =  342 

    342 = 2.3².19 = 18.19

    \(x\).(\(x+1\)) = 18.19

    \(x\) =   18

c, \(x\).(\(x\) + 1) = 650 

    650 = 2.5².13 = 25.26

    \(x\).(\(x\) +1) = 25.26

    \(x\) = 25

d, \(x\).(\(x\) +1) = 380 

    380 = 2².5.19 = 19.20

      \(x\).(\(x\) + 1) = 19.20

      \(x\)    = 19

43 + 2.(x - 342 ) = 123

2.(x-342 ) = 123 - 43

2.(x-342) = 80

x - 342 = 80 : 2

x - 342 = 40

x = 40 + 342

x = 382

ket qua la 382

120

120 : x, 360 : x và x >100

=> x là UCNN(120;360) và x< 100

120=2⁴ . 2⁵

360= 2³.3.5²

UCNN(120;360)=2³.3.5²= 600

=> U(600)={ 1;2;3;...600}

mà x > 100

=> x={ 100;.... 600}

Ta có: \(120⋮x\)

\(360⋮x\)

Do đó: \(x\inƯC\left(120;360\right)\)

\(\Leftrightarrow x\in\left\{1;2;3;4;5;6;8;10;12;15;20;24;30;40;60;120\right\}\)

mà x>100

nên x=120

Đáp án là C

Vì 162 ⋮ x; 360 ⋮ x ⇒ x ∈ ƯC (162, 360)

Ta có:

162 = 2. 3 4

360 = 2 3 . 3 2 .5

Suy ra, ƯCLN ( 162, 360) = 2. 3 2  = 18

ƯC ( 162, 360) = Ư (18) = {1, 2, 3, 6, 9, 18} ⇒ x ∈ {1, 2, 3, 6, 9, 18}

Vì 10 < x < 20 ⇒ x = 18