Số vô tỉ. Căn bậc hai

Vì $\sqrt{x}=\dfrac{3}{2}$ nên $x={{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}.$

Do đó: $2x=2.\dfrac{9}{4}=\dfrac{9}{2}.$

$\left( x-1 \right)-\dfrac{3}{2}=-\dfrac{1}{4}$

$x-1=-\dfrac{1}{4}+\dfrac{3}{2}$

$x-1=\dfrac{5}{4}$

$x=\dfrac{5}{4}+1$

$x=\dfrac{9}{4}.$

Khi đó: $\sqrt{x}=\sqrt{\dfrac{9}{4}}=\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}}=\dfrac{3}{2}.$

Vì $\sqrt{x}=5$ nên $x={{5}^{2}}=25.$

Do đó: $\,{{x}^{2}}={{25}^{2}}=625.$

$\dfrac{x}{3}=\dfrac{2}{x}\,$

$\Rightarrow \,x.x=3.2$

$\Rightarrow {{x}^{2}}=6$

$\Rightarrow x=\sqrt{6}$ hoặc $x=-\sqrt{6}$ .

$\sqrt{x-1}=\dfrac{2}{3}$

$\Rightarrow x-1={{\left( \dfrac{2}{3} \right)}^{2}}$

$\,\,\,\,\,\,x-1=\dfrac{4}{9}$

$\,\,\,\,\,x=\dfrac{4}{9}+1$

$\,\,\,\,\,x=\dfrac{13}{9}.$

${{2018}^{0}}.\sqrt{\dfrac{9}{49}}+{{\left( -\dfrac{1}{2} \right)}^{2}}.\sqrt{16}=1.\sqrt{{{\left( \dfrac{3}{7} \right)}^{2}}}+\dfrac{1}{4}.\sqrt{{{4}^{2}}}=\dfrac{3}{7}+1=\dfrac{10}{7}.$

Ta có:

$*\sqrt{36:4}=\sqrt{9}=\sqrt{{{3}^{2}}}=3\,;\,\,\sqrt{36}-\sqrt{4}=\sqrt{{{6}^{2}}}-\sqrt{{{2}^{2}}}=6-2=4$

$\Rightarrow \,\,\sqrt{36:4} < \sqrt{36}-\sqrt{4}$ .

$*\sqrt{25+16}=\sqrt{41}\,;\,$ $\sqrt{25}+\sqrt{16}=\sqrt{{{5}^{2}}}+\sqrt{{{4}^{2}}}=5+4=9=\sqrt{81}\, > \sqrt{41}$

$\Rightarrow \,\,\sqrt{25+16} < \sqrt{25}+\sqrt{16}$ .

$*\sqrt{25.9}=\sqrt{{{5}^{2}}{{.3}^{2}}}=15$ ; $\sqrt{25}.\sqrt{9}=\sqrt{{{5}^{2}}}.\sqrt{{{3}^{2}}}=5.3=15.$

$\Rightarrow \,\,\sqrt{25.9}=\sqrt{25}.\sqrt{9}$ .

$*\sqrt{\dfrac{25}{16}}=\sqrt{{{\left( \dfrac{5}{4} \right)}^{2}}}=\dfrac{5}{4} < \dfrac{5}{3}$ .

Ta có $\dfrac{74}{\sqrt{x}-1}$ là số nguyên khi $74\vdots \sqrt{x}-1$ $\left( x\ge 0 \right)$

Ta có bảng sau:

Vậy $x\in \left\{ 0,4,9,1444,5625 \right\}$ .

Gọi cạnh hình vuông là $x\left( cm \right),\,\,x > 0$

Diện tích hình vuông là: $S={{x}^{2}}\Rightarrow \,\,{{x}^{2}}=15\,\Rightarrow x=\sqrt{15}\approx 3,9$ (Vì $x > 0$ )

Vậy độ dài cạnh hình vuông là $3,9\,cm.$

$\sqrt{\dfrac{{{(-4)}^{2}}}{36}}=\sqrt{\dfrac{16}{36}}=\sqrt{\dfrac{4}{9}}=\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}}=\dfrac{2}{3};$

$\dfrac{\sqrt{36}-\sqrt{16}}{\sqrt{{{(-3)}^{2}}}}=\dfrac{\sqrt{{{6}^{2}}}-\sqrt{{{4}^{2}}}}{\sqrt{{{3}^{2}}}}=\dfrac{6-4}{3}=\dfrac{2}{3}\,;$

$\sqrt{\dfrac{{{(-5)}^{2}}-21}{3}}=\sqrt{\dfrac{25-21}{3}}=\sqrt{\dfrac{4}{3}}\,\,\ne \,\,\sqrt{\dfrac{4}{9}}=\,\dfrac{2}{3}$ ;

$\dfrac{\sqrt{25}-\sqrt{9}}{\sqrt{4}+1}=\dfrac{\sqrt{{{5}^{2}}}-\sqrt{{{3}^{2}}}}{\sqrt{{{2}^{2}}}+1}=\dfrac{5-3}{2+1}=\dfrac{2}{3}\,.$

Ta có $a\text{ }+\text{ }b$ là số vô tỉ nên $a\text{ }+\text{ }b$ là số hữu tỉ là sai.

$\sqrt{25}-\dfrac{1}{3}\sqrt{81}+\dfrac{3}{2}\sqrt{144}=\sqrt{{{5}^{2}}}-\dfrac{1}{3}\sqrt{{{9}^{2}}}+\dfrac{3}{2}\sqrt{{{12}^{2}}}=5-\dfrac{1}{3}.9+\dfrac{3}{2}.12=5-3+18=20.$

Nếu $x~\in \mathbb{Q}$ thì $x~\notin I$

Nếu $x~\in I$ thì $x~\notin \mathbb{Q}$

Nếu $x~\in \mathbb{R}$ thì $x~\in I$ hoặc $x~\in \mathbb{Q}$

Nếu $x\in \mathbb{Z}$ thì $x~\in \mathbb{R}$ .

Ta có $\sqrt{{{(-3)}^{4}}}-\sqrt{{{(-7)}^{2}}}+\sqrt{-{{(-4)}^{3}}}=\sqrt{{{3}^{4}}}-\sqrt{{{7}^{2}}}+\sqrt{{{4}^{3}}}=9-7+8=10$ .