Explicație pas cu pas:
1.a)
[tex]\left( \frac{1}{2} - \frac{1}{3}\right) \times \frac{9}{5} - 0.3 = \frac{3 - 2}{6} \times \frac{9}{5} - 0.3 \\ = \frac{1}{6} \times \frac{9}{5} - 0.3 = \frac{3}{10} - 0.3 = 0.3 - 0.3 = 0[/tex]
b)
[tex]\left(2 - \frac{1}{4}\right) \div \left(1 + 0.75\right) = \frac{8 - 1}{4} \div \left(1 + \frac{3}{4} \right) \\ = \frac{7}{4} \div \frac{7}{4} = 1[/tex]
2.
[tex]m_{a} = \frac{a + b + c}{3} = \frac{4 \sqrt{3} + 2 + \sqrt{48} - 4 + \frac{1}{2} + 3}{3} \\ = \frac{4 \sqrt{3} + 4 \sqrt{3} + 1 + \frac{1}{2} }{3} = \frac{8 \sqrt{3} + \frac{3}{2} }{6} = \frac{16 \sqrt{3} + 3}{6}[/tex]
3.
[tex]{x}^{2} + (m + 1)x + m + 1 = 0[/tex]
ecuația are două soluții reale => Δ > 0
[tex]{(m + 1)}^{2} - 4(m + 1) > 0 \\ (m + 1)(m + 1 - 4) > 0 \\ (m + 1)(m - 3) > 0 \\ m + 1 = 0\implies m = - 1 \\ m - 3 = 0\implies m = 3 \\ \iff m \in (- \infty ; -1) \cup (1; + \infty )[/tex]
4.
[tex]x \in \left( \frac{\pi}{2} ; \pi\right) \\ [/tex]
=> în cadranul II semnul: sinx +, cosx -
[tex]\sin^{2} ( \alpha ) + \cos^{2} ( \alpha ) = 1 \\ [/tex]
[tex]\cos^{2} (x) = 1 - \left(\frac{3}{5} \right)^{2} = 1 - \frac{9}{25} = \frac{16}{25} \\ [/tex]
[tex]\cos(x) = ± \frac{4}{5} \implies \cos(x) = - \frac{4}{5} \\ [/tex]
valoarea pozitivă nu aparține cadranului II
[tex]\cos(2x) = 2\cos^{2} (x) - 1 = \\ = 2 \cdot \frac{16}{25} - 1 = \frac{32 - 25}{25} = \frac{7}{25}[/tex]
[tex]\sin(2x) = 2 \sin(x) \cos(x) = \\ = 2 \cdot \frac{3}{5} \cdot \left( - \frac{4}{5} \right) = - \frac{24}{25} [/tex]
