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

         7.x + 2.x = 927

\(\Leftrightarrow\)( 7 + 2).x =927

\(\Leftrightarrow\)9.x = 927

\(\Leftrightarrow\)x = 927 : 9

\(\Leftrightarrow\)x = 103

Vậy x = 103

          3(x-2)+150=240

\(\Leftrightarrow\)3(x-2)=240-150

\(\Leftrightarrow\)3(x-2)=90

\(\Leftrightarrow\)x-2=90:3

\(\Leftrightarrow\)x-2=30

\(\Leftrightarrow\)x=30+2

\(\Leftrightarrow\)x=32

Vậy x=32

          ( x + 13 ) : 5 = 120

\(\Leftrightarrow\)x + 13 = 120 \(\times\)5

\(\Leftrightarrow\)x + 13 = 600

\(\Leftrightarrow\)x = 600 - 13

\(\Leftrightarrow\)x = 587

Vậy x = 587

          360:(x-7)=90

\(\Leftrightarrow\)x-7=360:90

\(\Leftrightarrow\)x-7=40

\(\Leftrightarrow\)x=40+7

\(\Leftrightarrow\)x=47

Vậy x=47

                                 7.x + 2.x = 927

                                 x.(7 + 2) = 927

                                 x.9 = 927

                                 x = 927 : 9

                                 x = 103

                             Vậy x = 103

                     3(x - 2) + 150 = 240

                    3(x - 2) = 240 - 150

                    3.(x - 2) = 90

                    x - 2 = 90 : 3 = 30

                    x = 30 + 2 = 32

                  Vậy x = 32

                 (x + 13) : 5 = 120

                 x + 13 = 120 x 5 = 600

                x = 600 - 13 = 587

                Vậy x = 587

              360:(x - 7) = 90

                    x - 7 = 360 : 90 = 4

                   x = 4 + 7 = 11

                   Vậy x = 11

             Ủng hộ mk nha !!! ^_^

a,[120-12*5]:6

    =[120-60]:6

    =60:6

     =10

b,494-[75+35:5]*2

=494-[75+7]*2

=494-82*2

=494-164

=330

c,1/4+[3/4:1/2]*5/6

=1/4+3/2*5/6

=1/4+5/4

=3/3

d,[360-15*6]:5

=[360-90]:5

=270:5

=54

a) (120-12*5):6= (120-60):6=60:6=10

b) 494-(75+35:5)*2= 494-(75+7)*2=494-82*2=494-164=330

c)1/4+(3/4:1/2)*5/6=1/4+3/2*5/6=1/4+5/4=6/4=3/2

d)(360-15*6):5=(360-90):5=270:5=54

a) ( 120 - 12 x 5 ) : 6

= ( 120 - 60 ) : 6

= 60 : 6

= 10

b) 494 - ( 75 + 35 : 5 ) x 2

= 494 - ( 75 + 7 ) x 2

= 494 - 82 x 2

= 494 - 164

= 330

c) 1/4 + ( 3/4 : 1/2 ) x 5/6

= 1/4 + 3/2 x 5/6

= 1/4 + 5/4

= 6/4 = 3/2

d) ( 360 - 15 x 6 ) : 5

= ( 360 - 90 ) : 5

= 270 : 5

= 54

Bài 1:

a: UCLN(150;1200)=150

UC(150;1200)=Ư(150)={1;2;3;5;6;10;15;25;30;50;75;150}

b: UCLN(120;160;40)=40

ƯC(120;160;40)=Ư(40)={1;2;4;5;8;10;20;40}

Bài 2: 

\(\Leftrightarrow x\inƯC\left(360;930;450\right)\)

\(\Leftrightarrow x\inƯ\left(90\right)\)

mà x>30

nên \(x\in\left\{45;90\right\}\)

\(12\left(x+5\right)+2x=130\\\Leftrightarrow 12x+60+2x=130\\ \Leftrightarrow14x=70\\ \Leftrightarrow x=5\\ ----\\ 23\left(x-5\right)-12x=138\\ \Leftrightarrow23x-115-12x=138\\ \Leftrightarrow23x-12x=138+115\\ \Leftrightarrow11x=253\\ \Leftrightarrow x=\dfrac{253}{11}=23\\ ----\\ 360-12x+23\left(x-5\right)=278\\ \Leftrightarrow360-12x+23x-115=278\\ \Leftrightarrow-12x+23x=278+115-360\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=\dfrac{33}{11}=3\)

\(6\left(x+3\right)+3\left(x-5\right)=278\\ \Leftrightarrow6x+18-3x-15=278\\ \Leftrightarrow6x-3x=278+15-18\\ \Leftrightarrow3x=275\\ \Leftrightarrow x=\dfrac{275}{3}\\ ---\\ \left(7-x\right)\left(3x-90\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-90=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=30\end{matrix}\right.\)

\(\left(x-2\right)\left(4x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\\ ---\\ \left(2x-18\right)\left(3x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-18=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\3x=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)

a, 70=2.5.10; 90=2.3².5

=> ƯCLN(70;90)=2.5=10 => ƯC(70;90)=Ư(10)={1;2;5;10}

b, 180=2².3².5 ; 235= 47.5; 120=2³.3.5

=> ƯCLN(180;235;120)= 5 => ƯC(180;235;120)=Ư(5)={1;5}

Mình xét ước tự nhiên thui ha

Trên là bài 1, dưới này là bài 2!

a, 480 và 720 đều chia hết cho x

480=2⁵.3.5; 720= 2⁴.3².5

=> ƯCLN(480;720)=2⁴.3.5=240

=> x=ƯCLN(480;720)=240

b, 240 và 360 đều chia hết cho x

240=2⁴.3.5; 360=2³.3².5

=>ƯCLN(240;360)=2³.3.5=120

x=ƯCLN(240;360)=120

C1:\(a,360:\left(3.5\right)\)

\(=360:15\)

\(=24\)

C2:\(\left(360:3\right):5\)

\(=120:5\)

\(=24\)

C1:\(b,1440:\left(2.5\right)\)

\(=1440:10\)

\(=144\)

C2:\(\left(1440:2\right):5\)

\(=720:5\)

\(=144\)

\(a,C1:360:\left(3\times5\right)=360:15=24\\ C2:360:\left(3\times5\right)=\left(360:3\right):5=120:5=24\\ b,C1:1440:\left(2\times5\right)=1440:10=144\\ C2;1440:\left(2\times5\right)=\left(1440:2\right):5=720:5=144\)

C1:

C2:

C1:

ĐẶT \(\frac{x}{2}=\frac{y}{5}=k\)

=>\(\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)

Ta có xy=360

(=)2k.5k=360

(=)10\(k^2\)=360

(=)\(k^2=36\)

=>\(\orbr{\begin{cases}k=6\\k=-6\end{cases}}\)

*Với k=6=>\(\hept{\begin{cases}x=12\\y=30\end{cases}}\)

*Với k=-6=>\(\hept{\begin{cases}x=-12\\y=-30\end{cases}}\)