a) x - 3 = -6

x=(-6)+3

x=-3

b) x - (-5)=4

x=4+5

x=9

c)x - (-9) = 4-(-9)

x - (-9) =13

x =13-9

x = 4

d) 4-x=-3-(-6)

4-x=3

x=4-1

x=3

a) -3

b) 9

c)4

d)3

a)4/9 : x/6 = 2/3

           x/6 = 4/9 : 2/3

          x/6  =   2/3 = 4/6

vậy x = 4

b)x/5 x 3/7 = 9/35

   x/5          = 9/35 : 3/7

   x/5          = 3/5

vậy x = 3

a. = -3

b. = -1

c. = 4

d. = 1

a)x=-3

b)x=-1

c)x=4

d)x=1

Trả lời 

\(x.\left(x+2\right).\left(x+4\right).\left(x+6\right)=9\)

\(\Leftrightarrow\left[x.\left(x+6\right)\right].\left[\left(x+2\right).\left(x+4\right)\right]=9\)

\(\Leftrightarrow\left(x^2+6x\right).\left(x^2+6x+8\right)=9\)

Đặt \(x^2+6x=t\) ta có 

\(t.\left(t+8\right)=9\)

\(\Leftrightarrow t^2+8t-9=0\)

\(\Leftrightarrow t^2-t+9t-9=0\)

\(\Leftrightarrow t.\left(t-1\right)+9.\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right).\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t+9=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=1\\t=-9\end{cases}}\)

TH1 \(t=1\)

\(\Rightarrow x^2+6x=1\)

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

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

\(\Leftrightarrow\left(x+3\right)^2=10=\left(\pm\sqrt{10}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3+\sqrt{10}\\x=-3-\sqrt{10}\end{cases}}\)

TH2: \(t=-9\)

\(\Rightarrow x^2+6x=-9\)

\(\Leftrightarrow x^2+6x+9=0\)

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

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy \(x\in\left\{-3+\sqrt{10};-3-\sqrt{10};-3\right\}\)

\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x-9=0\)

\(\Leftrightarrow\left(x^3+9x^2+17x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)

TH1 : Ta có : \(6^2-4.\left(-1\right)=36+4=40>0\)Suy ra : \(x_1=\frac{-6-\sqrt{40}}{2};x_2=\frac{-6+\sqrt{40}}{2}\)

TH2 : \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

\(\left(\dfrac{4}{9}\right)^{x+1}=\left(\dfrac{8}{27}\right)^6\)

\(\Leftrightarrow\left(\dfrac{2}{3}\right)^{2x+2}=\left(\dfrac{2}{3}\right)^{18}\)

\(\Leftrightarrow2x+2=18\Leftrightarrow2x=16\Leftrightarrow x=8\)

+) X-3=-6

=> x = -6 +3

=> x = -3 

+) x-(-5) =4

=> x = 4 + (-5)

=> x = -1

+) X-(-9) =4-(-9)   

=> x =  4 - (-9) + (-9)

=> x = 4

+) 4-x=-3-(-6) 

=> 4-x = 3

=> x = 4-3 

=> x = 1