Gọi q là thương phép chia a cho 6 ta có:

\(a=6q+2\)

\(\Rightarrow a^2=\left(6q+2\right)^2\Leftrightarrow a^2=36q^2+24q+4\\ \Leftrightarrow a^2=6\left(6q^2+4q\right)+4\)

Vậy \(a^2\) chia cho 6 được thương là \(6q^2+4q\) và dư 4 . đpcm

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)

\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\end{matrix}\right.\)

.

\(\dfrac{x^2.30+y^2.20+z^2.15}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\\ \Rightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\\ \Rightarrow30x^2-12x^2+20y^2-12y^2+15z^2-12z^2=0\\ \Rightarrow18x^2+8y^2+3z^2=0\\ \)

Mà \(x^2\ge0\forall x\\ y^2\ge0\forall y\\ z^2\ge0\forall z\)

=> \(\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

.

a) x²-5x+4

= x²-x-(4x-4)

= x(x-1)-4(x-1)

= (x-4)(x-1)

b) giống phần a

c) x²-7x+12

= x²-3x-(4x-12)

= x(x-3)-4(x-3)

= (x-4)(x-3)

d) x²+7x+12

= x²+3x+4x+12

= x(x+3)+4(x+3)

= (x+3)(x+4)

e) x²-8x+12

= x²-2x-(6x-12)

= x(x-2)-6(x-2)

=(x-6)(x-2)

f)x²+8x+12

= x²+2x+6x+12

=x(x+2)+6(x+2)

=(x+2)(x+6)

g) x²+x-12

=x²-3x+4x-12

=x(x-3)+4(x-3)

=(x-3)(x+4)

h)x²-3x-10

=x²+2x-(5x+10)

=x(x+2)-5(x+2)

=(x-5)(x+2)

i)x²+4x-21

=x²-3x+7x-21

=x(x-3)+7(x-3)

=(x+7)(x-3)

Mn giúp mình với ạ, mình đg cần gấp để ôn thi ạ.