\(a,\dfrac{x}{7}=\dfrac{6}{21}\Rightarrow x.21=7.6=42\\ \Rightarrow x=2\\ b,\dfrac{1}{2}=\dfrac{x}{12}\Rightarrow1.12=2.x=12\\ \Rightarrow x=6\\ c,\dfrac{-7}{6}=\dfrac{x}{12}\Rightarrow-7.12=6.x=-84\\ \Rightarrow x=-14\)

mình đang cần gấp, giúp mình với ạ

a ,   6 × x = 1 2 x = 1 2 : 6 x = 1 12 b ,   x : 3 7 = 7 11 x = 7 11 × 3 7 x = 3 11

a: \(\left(-11\right)\cdot\left(-28\right)+\left(-9\right)\cdot13\)

\(=308-117\)

=191

b: \(\left(-69\right)\cdot\left(-31\right)-\left(-15\right)\cdot12\)

\(=2139-\left(-180\right)\)

=2139+180

=2319

c: \(\left[16-\left(-5\right)\right]\cdot\left(-7\right)\)

\(=\left(16+5\right)\cdot\left(-7\right)\)

\(=-7\cdot21=-147\)

d: \(\left[\left(-4\right)\cdot\left(-9\right)-6\right]\left[\left(-12\right)-\left(-7\right)\right]\)

\(=\left[36-6\right]\left[-12+7\right]\)

\(=30\cdot\left(-5\right)=-150\)

`a)(x-6)^2-(x+6)^2=12`

`<=>(x-6-x-6)(x-6+x+6)=12`

`<=>-12.2x=12`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2`

`b)36x^2-12x+1=81`

`<=>(6x-1)^2=81`

`<=>(6x-1-9)(6x-1+9)=0`

`<=>(6x-10)(6x+8)=0`

`<=>(3x-5)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x-6)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2-6x+x-6=0`

`<=>x(x-6)+x-6=0`

`<=>(x-6)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

Này bn, bn đừng tưởng thế. bn bt giải thì hộ mik cái mik mới vào lớp 7. đã lên trên mạng rồi mà còn chửi vs nhau không có đạo đức à

bn ko trả lời lại ak

a: -29-(15-x)=34

=>-29-15+x=34

=>x-44=34

=>x=34+44

=>x=78

b: 15-(x-17)=-12

=>x-17=15+12=27

=>x=27+17

=>x=44

c: -(x-6+85)=(x+51)-54

=>-(x+79)=x+51-54

=>x-3=-x-79

=>2x=-79+3=-76

=>x=-38

a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Ai giúp vs

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

a)  × 4 = 28

28 : 4 = 7

Vậy 7 × 4 = 28.

b) × 3 = 12

12 : 3 = 4

Vậy 4 × 3 = 12.

c) 6 ×  = 24

24 : 6 = 4

Vậy 6 × 4 = 24.