a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

\(-\dfrac{8}{11}.x=\dfrac{2}{5}.\dfrac{1}{4}\)

\(-\dfrac{8}{11}.x=\dfrac{2}{20}\)

\(-\dfrac{8}{11}.x=\dfrac{1}{10}\)

\(x=\dfrac{1}{10}:-\dfrac{8}{11}\)

\(x=-\dfrac{11}{81}\)

Chọn d

D nha

a) 

B(14) = 0; 14; 28; 42; 56; 70; 84; …..

Vì 20 < x < 80 => x ∈ { 28; 42; 56; 70．}

b)

Vì 70 chia hết cho x ѵà 80 chia hết cho x => x ∈ ƯC(70; 80)

Phân tích:

70 = 2 ．5 ．7

80 = 24 ．5

ƯCLN (70; 80) = 2.5=10

ƯC ( 70; 80) = Ư(10) ={1;2;5;10}

Mà x > 8 => x = 10

c)

Vì 126 chia hết cho x ѵà 210 chia hết cho x => x ∈ ƯC(126; 210)

Phân tích

126 = 2 ．3² ．7

210 = 2 ．3 ．5 ．7

ƯCLN(126; 210) = 2 ．3 ．7 = 42

ƯC(126; 210) = { 1; 2; 3; 6; 7; 14; 21; 42 }

Vì 15 < x < 30 => x = 21

a) \(B\left(14\right)=\left\{0;14;28;42;56;70;84;.......\right\}\)

Vì \(20< x< 8\Rightarrow x\in\left\{28;42;56;70\right\}\)

b) Vì 70 chia hết cho x và 80 chia hết cho x nên \(\Rightarrow x\inƯC\left(70;80\right)\)

Phân tích :

\(70=2.5.7\)

\(80=2^4.5\)

Mà \(x>8\Rightarrow x=10\)

c) Vì 126 chia hết cho x  và 210 chia hết cho x nên \(\Rightarrow x\inƯC\left(126;210\right)\)

Phân tích :

\(126=2.3^2.7\)

\(210=2.3.5.7\)

\(ƯC\left(126;210\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

\(ƯCLN\left(126;210\right)=2.3.7=42\)

Theo đề : \(x\in\left(>15< 30\right)\Rightarrow x=21\)

a. \(\dfrac{x}{5}=\dfrac{8}{10}\)

\(\Rightarrow10x=8.5\)

\(\Rightarrow10x=40\)

\(\Rightarrow x=\dfrac{40}{10}=4\)

b. \(\dfrac{9}{x}=\dfrac{90}{100}\)

\(\Rightarrow9.100=90x\)

\(\Rightarrow900=90x\)

\(\Rightarrow x=\dfrac{900}{90}=10\)

c. \(17+\dfrac{x}{35}=\dfrac{5}{7}\)

\(\Rightarrow\dfrac{595+x}{35}=\dfrac{25}{35}\)

\(\Rightarrow595+x=25\)

\(\Rightarrow x=25-595=-570\)

d. \(x:\dfrac{5}{27}=\dfrac{2}{9}\)

\(\Rightarrow\dfrac{27x:5}{27}=\dfrac{6}{27}\)

\(\Rightarrow27x:5=6\)

\(\Rightarrow27x=5.6\)

\(\Rightarrow27x=30\)

\(\Rightarrow x=\dfrac{30}{27}=\dfrac{10}{9}\approx1,11\)

`a)(x-2)(x^2+2x+4)-x(x-3)(x+3)=26`

`<=>x^3-2^3-x(x^2-9)=26`

`<=>x^3-8-x^3+9x=26`

`<=>9x-8=26`

`<=>9x=34`

`<=>x=34/9`

`b)(x-3)(x^2+3x+9)-x(x+4)(x-4)=21`

`<=>x^3-3^3-x(x^2-16)=21`

`<=>x^3-27-x^3+16x=21`

`<=>16x-27=21`

`<=>16x=48`

`<=>x=3`

1.64a=80b=96c=>\(\frac{64a}{960}=\frac{80b}{960}=\frac{96c}{960}\)

=>\(\frac{a}{15}=\frac{b}{12}=\frac{c}{10}\)

......ko biết

2.Có:xy+3x+y=4

=>x(y+3)+y=4

=>x(y+3)+(y+3)=4+3=7

=>(x+1)(y+3)=7=>x+1 và y+3 thuộc Ư(7)

Với các cặp số(x;y) trên ko có số nào thỏa mãn x+y=19

Ta có:     64=2.2.2.2.2.2

               80=2.2.2.2.5

               96=2.2.2.2.2.3

=>BCLN(64,80,96)=2.2.2.2.2.2.3.5=960

Vì a,b,c nhỏ nhất nên 64a=80b=96c

=>a=960:64=15

    b=960:80=12

    c=960:96=10

                              Vậy a=15 ; b=12 ; c=10

Thay BCLN thành BCNN

a) (y – 1) : 105 = 125 x 80

(y – 1) : 105 = 10000

(y – 1) = 10000 x 105

(y – 1) = 1050000

y = 1050000 + 1

y = 1050001

b) (y – 607 200) : 305 = 642 + 318

(y – 607 200) : 305 = 960

(y – 607 200) = 960 x 305

(y – 607 200) = 292800

y = 292800 + 607 200

y = 900000

\(a.\left(y-1\right):105=125x80\)                   \(b.\left(y-607200\right):305=642+318\)

\(\left(y-1\right):105=10000\)                             \(\left(y-607200\right):305=960\)

\(y-1=10000x105=1050000\)                \(y=960+607200=608160\)

\(y=1050000+1=1050001\)

\(c.\left(1780-973\right)x\left(75:y\right)=2401+20\)

\(807x\left(75:y\right)=2401+20\)

\(807x\left(75:y\right)=2421\)

\(75:y=2421:807=3\)

\(y=75:3=25\)

a)  (y - 1) : 105 = 125 x 80

     (y - 1) : 105 = 10000

     (y - 1)          = 10000 x 105

     (y - 1)         = 10500000

      y                = 10500000 + 1

      y                = 10500001.

b) (y – 607 200) : 305 = 642 + 318

(y – 607 200) : 305 = 960

(y – 607 200) = 960 x 305

(y – 607 200) = 292800

 y = 292800 + 607 200

 y = 900000.

       y = 25.

giúp mik với cảm ơn 

a) \(\left|x-17\right|=2,3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-17=2,3\\x-17=-2,3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=19,3\\x=14,7\end{matrix}\right.\)

b) \(\left|x+\dfrac{3}{4}\right|=0\)

\(\Leftrightarrow x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{4}\)

c) \(\left|x+\dfrac{3}{4}\right|+\dfrac{1}{3}=0\)

\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=-\dfrac{1}{3}\)( vô lý do \(\left|x+\dfrac{3}{4}\right|\ge0\forall x\))

Vậy \(S=\varnothing\)

a) \(\left(x-3\right)^2+\left(4-x\right)\left(x+4\right)=10\)

\(\Leftrightarrow\left(x^2-2\cdot x\cdot3+3^2\right)+\left(4-x\right)\left(4+x\right)=10\)

\(\Leftrightarrow x^2-6x+9+\left(4^2-x^2\right)-10=0\)

\(\Leftrightarrow x^2-6x-1+16-x^2=0\)

\(\Leftrightarrow-6x+15=0\)

\(\Leftrightarrow6x=15\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

b) \(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c) \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x^2-3^2\right)^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)

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