`x^2-3x+5=x^2-2x. 3/2+9/4+11/4=(x-3/2)^2+11/4`

Vì \((x-3/2)^2 \ge 0\)

\(<=>(x-3/2)^2+11/4 \ge 11/4\)

\(<=>x^2-3x+5 \ge 11/4\)

\(<=>5:[x^2-3x+5] \le 20/11\)

Dấu "`=`" xảy ra `<=>(x-3/2)^2=0<=>x=3/2`

Ta có: `x^2-3x=x^2-2x. 3/2+9/4-9/4=(x-3/2)^2-9/4`

 Vì \((x-3/2)^2 \ge 0\)

\(<=>(x-3/2)^2-9/4 \ge -9/4\)

\(<=>x^2-3x \ge -9/4\)

\(<=>5/[x^2-2x] \le -20/9\)

Dấu "`=`" xảy ra `<=>(x-3/2)^2=0<=>x=3/2`

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

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

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