ƯCLN(90 , 126 ) = 9

ƯC(90,26) = Ư(9) = 1,3,9

\(90=2\cdot3^2\cdot5\\ 126=2\cdot3^2\cdot7\\ ƯCLN\left(90,126\right)=2\cdot3^2=18\\ ƯC\left(90,126\right)=Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

\(\dfrac{5}{7}\)

\(\dfrac{90}{126}=\dfrac{5}{7}\)

P=x2−4x+5

  =x2−4x+4+1

  =(x−2)2+1≥1

⇒Pmin=1 khi x=2

a: Sửa đề: S=5+5^2+...+5^2006

5S=5^2+5^3+...+5^2007

=>4S=5^2007-5

=>S=(5^2007-5)/4

b: S=5+5^4+5^2+5^5+...+5^2003+5^2006

=5(1+5^3)+5^2(1+5^3)+...+5^2003(1+5^3)

=126(5+5^2+...+5^2003) chia hết cho 126

Để pt có 2 nghiệm thì \(\Delta'=m^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge2\\m\le-2\end{matrix}\right.\).

Khi đó theo hệ thức Viète ta có \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=4\end{matrix}\right.\).

Ta có \(\left(x_1+1\right)^2+\left(x_2+1\right)^2=2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)=0\)

\(\Leftrightarrow\left(2m\right)^2-2.4+2.2m=0\Leftrightarrow m^2+m-2=0\Leftrightarrow\left(m-1\right)\left(m+2\right)=0\Leftrightarrow\left[{}\begin{matrix}m=1\left(l\right)\\m=-2\left(TM\right)\end{matrix}\right.\).

Vậy m = -2.

Mn ơi giúp mình với ạ❤

\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)

\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)