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

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

\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)

\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)

\(=x^2+6x+9-17=\left(x+3\right)^2-17=\left(x+3-\sqrt{17}\right)\left(x+3+\sqrt{17}\right)\)

\(9^{10}-3^{18}\\ =\left(3^2\right)^{10}-3^{18}=3^{20}-3^{18}\\ =3^{18}\left(3^2-1\right)\\ =3^{18}\left(9-1\right)\\ =3^{18}.8⋮8\)

\(\dfrac{x}{7}=\dfrac{18}{14}\)

<=> 14 . x = 18 . 7

<=> 14x = 126

<=> x = \(\dfrac{126}{14}=9\)

40/12

y = 8 : 12/5

y =

y = 8 : 12/5

y = 12 / 40

x³-25x=0

 x(x²-25)=0

x(x²-5²)=0

x(x-5)(x+5)=0

=> x=0 ; x-5= 0 hoặc x+5= 0

=> x=0; x=5 hoặc x= -5

=> x ={ 0;5;-5}

x^3 - 25x = 0
<=> x( x^2 - 25) = 0
<=> x( x - 5)(x + 5) = 0
<=> x = 0
x -5 = 0
x + 5 = 0
<=> x = 0
x = 5
x = -5

a: Tọa độ tâm là:

x=(6+0)/2=3 và y=(0+8)/2=4

\(IA=\sqrt{\left(3-6\right)^2+\left(4-0\right)^2}=5\)

=>(C): (x-3)^2+(y-4)^2=25

bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)

\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)

\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)

\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)

\(=\dfrac{-27.1.1}{5.1.9}\)

\(=\dfrac{-27}{45}\)

\(=\dfrac{-9}{15}\)

b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)

\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)

\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)

\(=8+3.1-\dfrac{1}{4}.4+8.8\)

\(=8+3-1+64\)

\(=11-1+64\)

\(=10+64\)

\(=74\)