Question

10 КЛАСС ПОМОГИТЕ С ПРИМЕРАМИ ПЛИЗ✌ 70 баллов!!

Answer 1

1) α - угол третьей четверти, значит Cosα < 0 , tgα > 0 .

$Cos\alpha=-\sqrt{1-Sin^{2}\alpha} =-\sqrt{1-(-0,2)^{2} }=-\sqrt{1-0,04} =-\sqrt{0,96}=\\\\=-\dfrac{4\sqrt{6} }{10}=-\dfrac{2\sqrt{6} }{5}=-0,4\sqrt{6} \\\\tg\alpha=\dfrac{Sin\alpha }{Cos\alpha }=\dfrac{-0,2}{-0,4\sqrt{6} }=\dfrac{\sqrt{6} }{12} \\\\\\2) \ x^{2}-6x+8\leq 0\\\\(x-2)(x-4)\leq 0$

+ + + + + [2] - - - - -[4] + + + + +

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Ответ : x ∈ [2  ;  4]

$3) \ \sqrt[3]{3x^{2}-1 } =3\\\\(\sqrt[3]{3x^{2}-1 })^{3} =3^{3} \\\\3x^{2}-1=27\\\\3x^{2}=28\\\\x^{2}=\dfrac{28}{3} \\\\x_{1}=\sqrt{9\frac{1}{3} } \ ; \ x_{2}=-\sqrt{9\frac{1}{3} }\\\\\\4) \ \Big(\dfrac{1}{3}\Big)^{4x-1}=3\\\\\Big(\dfrac{1}{3}\Big)^{4x-1}=\Big(\dfrac{1}{3}\Big)^{-1} \\\\4x-1=-1\\\\4x=0\\\\x=0$

$5) \ 3\sqrt[3]{0,125}-2\sqrt[3]{0,064}=3\cdot\sqrt[3]{0,5^{3} } -2\cdot\sqrt[3]{0,4^{3} }=3\cdot0,5-2\cdot0,4=\\\\=1,5-0,8=0,7$