Explicație pas cu pas:
11.
[tex] \sqrt{4 \sqrt{25} + {5}^{2} \cdot7 } = \sqrt{4\cdot 5 + 25 \cdot7 } = \sqrt{20 + 175} \\ = \sqrt{195} = \sqrt{3\cdot5\cdot13} [/tex]
[tex]\sqrt{ {3}^{3}\cdot 4 - {3}^{3} + 9 \sqrt{9} } = \sqrt{ {3}^{3}\cdot {2}^{2} - {3}^{3} + {3}^{2}\cdot 3} \\ = \sqrt{ {3}^{3}\cdot {2}^{2} - {3}^{3} + {3}^{3}} = \sqrt{ {3}^{3}\cdot {2}^{2} } = 2 \cdot 3 \sqrt{3} = 6 \sqrt{3} [/tex]
[tex] \sqrt{ \sqrt{225} + 6 \sqrt{36} + {5}^{3}} = \sqrt{15 + 6 \cdot 6 + 125} \\ = \sqrt{140 + 36} = \sqrt{176} = \sqrt{ {2}^{4}\cdot 11 } \\ = {2}^{2} \sqrt{11} = 4 \sqrt{11}[/tex]
12.a)
[tex] \sqrt{12 \cdot x} = \sqrt{ {2}^{2}\cdot 3 \cdot x} = 2 \sqrt{3\cdot x} \\ = > x = 3[/tex]
12.b)
[tex]\sqrt{48 \cdot x} = \sqrt{ {2}^{4}\cdot 3 \cdot x} = {2}^{2} \sqrt{3\cdot x} \\ = > x = 3[/tex]
1.a)
[tex]P = 4 \cdot l = 4 \sqrt{5}\: cm \\ A = {l}^{2} = {( \sqrt{5} )}^{2} = 5 \: {cm}^{2} [/tex]
1.b)
[tex]P = 3 \cdot l = 3 \cdot 2 \sqrt{3} = 6 \sqrt{3} \: cm \\ A = \frac{ {l}^{2} \sqrt{3}}{4} = \frac{({2 \sqrt{3}) }^{2}\cdot \sqrt{3} }{4} \\ = \frac{12\cdot \sqrt{3} }{4} =3 \sqrt{3} \: {dm}^{2} [/tex]
