ВПР 2 №35 Алгебра = ВПТ 7 №35 Математика
Обчисліть:
1. $\sqrt{4\frac{1}{20}}\cdot\sqrt{2\frac{2}{9}}-(\sqrt7)^2;$
2. $\sqrt{1-\left(\frac{8}{17}\right)^2}+\left(-\sqrt{\frac{2}{17}}\right)^2;$
3. $\sqrt{1-\left(-\frac{3}{5}\right)^2}-\left(-\frac{\sqrt2}{\sqrt5}\right)^2;$
4. $\sqrt{2\frac{1}{5}}\cdot\sqrt{1\frac{1}{11}}\cdot\sqrt{2\frac{2}{5}}+$
$+ \left(\sqrt{\frac{3}{5}}\right)^2-(\sqrt3)^2.$
Розв'язок:
1. $\sqrt{4\frac{1}{20}}\cdot\sqrt{2\frac{2}{9}}-(\sqrt7)^2=$
$= \sqrt{\frac{81}{20}\cdot\frac{20}{9}}-7=$
$= \sqrt{\frac{81}{20}\cdot\frac{20}{9}}-7=$
$= \sqrt9-7=3-7=-4;$
2. $\sqrt{1-\left(\frac{8}{17}\right)^2}+\left(-\sqrt{\frac{2}{17}}\right)^2=$
$= \sqrt{\left(1-\frac{8}{17}\right)\left(1+\frac{8}{17}\right)}+\frac{2}{17}=$
$= \sqrt{\frac{9}{17}\cdot\frac{25}{17}}+\frac{2}{17}=$
$= \frac{\sqrt{9\cdot25}}{\sqrt{{17}^2}}+\frac{2}{17}=$
$= \frac{3\cdot5}{17}+\frac{2}{17}=\frac{15}{17}+\frac{2}{17}=$
$= \frac{17}{17}=1.$
3. $\sqrt{1-\left(-\frac{3}{5}\right)^2}-\left(-\frac{\sqrt2}{\sqrt5}\right)^2=$
$= \sqrt{1-\frac{9}{25}}-\frac{2}{5}=$
$= \sqrt{\frac{25}{25}-\frac{9}{25}}-\frac{2}{5}=$
$= \sqrt{\frac{16}{25}}-\frac{2}{5}=$
$= \frac{4}{5}-\frac{2}{5}=0,4;$
4. $\sqrt{2\frac{1}{5}}\cdot\sqrt{1\frac{1}{11}}\cdot\sqrt{2\frac{2}{5}}+$
$+ \left(\sqrt{\frac{3}{5}}\right)^2-(\sqrt3)^2=$
$= \sqrt{\frac{11}{5}}\cdot\sqrt{\frac{12}{11}}\cdot\sqrt{\frac{12}{5}}+$
$+ \frac{3}{5}-3= \sqrt{\frac{11\cdot12\cdot12}{5\cdot11\cdot5}}+\frac{3}{5}-3=$
$= \sqrt{\frac{{12}^2}{5^2}}+\frac{3}{5}-3=$
$= \frac{12}{5}+\frac{3}{5}-3=$
$= \frac{15}{5}-3=3-3=0.$