§3. Các phép toán tập hợp

ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-1}=\sqrt{x-2}+\sqrt{x-3}\)

\(\Leftrightarrow x-1=x-2+x-3+2\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow2\sqrt{x^2-5x+6}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-5x+6\right)=\left(4-x\right)^2\\x\le4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-12x+8=0\\x\le4\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{6+2\sqrt{3}}{3}\left(tm\right)\)

ĐK: \(\begin{cases}x≥1\\x≥2\\x≥3\end{cases} \Leftrightarrow x≥3\)

\(\sqrt{x-1}-\sqrt{x-2}=\sqrt{x-3} \\ \Leftrightarrow \sqrt{x-1}=\sqrt{x-3}+\sqrt{x-2} \\ \Leftrightarrow x-1= x-3+x-2+2\sqrt{(x-2)(x-3)} \\\Leftrightarrow -x+4=2\sqrt{(x-2)(x-3)} \\ \Leftrightarrow x^2-8x+16=4(x-2)(x-3) (ĐK: 3≤x≤4) \\ \Leftrightarrow x^2-8x+16=4x^2-20x+24\\\Leftrightarrow3x^2-12x+8=0 \\ \Leftrightarrow x = \dfrac{6 \pm 2\sqrt3}{3}\)

Vậy \(x=\dfrac{6+2\sqrt3}{3}\).

\(\left\{{}\begin{matrix}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\\\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-\overrightarrow{AB}+\overrightarrow{AD}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\frac{\overrightarrow{AC}-\overrightarrow{BD}}{2}\\\overrightarrow{AD}=\frac{\overrightarrow{AC}+\overrightarrow{BD}}{2}\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AD}=\left(\frac{\overrightarrow{AC}-\overrightarrow{BD}}{2}\right)\left(\frac{\overrightarrow{AC}+\overrightarrow{BD}}{2}\right)=\frac{\overrightarrow{AC}^2-\overrightarrow{BD}^2}{4}=\frac{3^2-7^2}{4}=-10\)

\(\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}+\sqrt{3}\left(x^2-3x+1\right)=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow ab+\sqrt{3}\left(2a^2-b^2\right)=0\)

\(\Leftrightarrow\left(\sqrt{3}a+b\right)\left(2a-\sqrt{3}b\right)=0\)

\(\Leftrightarrow4a^2=3b^2\)

\(\Leftrightarrow4\left(x^2-x+1\right)=3\left(x^2+x+1\right)\)

\(\Leftrightarrow...\)

\(\overline{52ab}\) hay là 52 \(\times\)ab

Lời giải:

Vì $x+y+z=0\Rightarrow x+y=-z$.

Ta có:

\(A=x^3+y^3+x^2z+y^2z-xyz+2020^{10}\)

\(=(x+y)(x^2-xy+y^2)+z(x^2+y^2)-xyz+2020^{10}\)

\(=-z(x^2-xy+y^2)+z(x^2+y^2)-xyz+2020^{10}=-z(x^2+y^2)+xyz+z(x^2+y^2)-xyz+2020^{10}\)

\(=2020^{10}\)

Lời giải:

$P=4x^2+2y^2+4xy-4x+2y+2025=(4x^2+4xy+y^2)+y^2-4x+2y+2025$

$=(2x+y)^2-2(2x+y)+y^2+4y+2025$

$=(2x+y)^2-2(2x+y)+1+(y^2+4y+4)+2020$

$=(2x+y-1)^2+(y+2)^2+2020$

$\geq 2020$

Vậy GTNN của $P$ là $2020$. Giá trị này đạt tại \(\left\{\begin{matrix} 2x+y-1=0\\ y+2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{3}{2}\\ y=-2\end{matrix}\right.\)

Xét phép chia \(5x^3+4x^2-6x-a\) cho \(5x-1\)

Phép chia là chia hết khi \(-a-1=0\Leftrightarrow a=-1\)