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7)\(\frac{1}{1-x^2}>\frac{3x}{\sqrt{1-x^2}}-1\)(-1<x<1)
Đặt a=1-x2 ta được: (ĐK a>0)
\(\frac{1}{a}>\frac{3x}{\sqrt{a}}-1\)
\(\Leftrightarrow\frac{1}{a}-\frac{3\sqrt{a}x}{a}+\frac{a}{a}>0\)
\(\Leftrightarrow\frac{1-3\sqrt{a}x+a}{a}>0\)
\(\Leftrightarrow1-3\sqrt{a}x+a>0\left(a>0\right)\)
\(\Leftrightarrow1-3\sqrt{x^2-1}.x+x^2-1>0\)
\(\Leftrightarrow x^2>3\sqrt{x^2-1}x\)
<=>x4 > 9.(x2-1).x2
<=>x4>9x4-9x2
<=>8x4-9x2<0
<=>x2.(8x2-9)<0
<=>8x2-9<0
<=>x2<9/8
=>\(-\frac{3\sqrt{2}}{4}\)<x<\(\frac{3\sqrt{2}}{4}\)
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\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-7}{156}\)
\(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-17}{12}\)
\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-7}{55}\)
\(\dfrac{-34}{37}.\dfrac{74}{-85}=\dfrac{4}{5}\)
\(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{10}{7}\)
Chúc bạn học tốt!!!
a) \(\left(-\dfrac{1}{39}\right)+\left(-\dfrac{1}{52}\right)=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
b) \(\left(-\dfrac{6}{9}\right)+\left(-\dfrac{12}{16}\right)=-\dfrac{6}{9}-\dfrac{12}{16}=-\dfrac{17}{12}\)
c) \(-\dfrac{2}{5}-\left(-\dfrac{3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}=-\dfrac{7}{55}\)
d) \(\left(-\dfrac{34}{37}\right)\cdot\left(-\dfrac{74}{85}\right)=2\cdot\dfrac{2}{5}=\dfrac{4}{5}\)
e) \(\left(-\dfrac{5}{9}\right):\left(-\dfrac{7}{18}\right)=\dfrac{5}{9}\cdot\dfrac{18}{7}=5\cdot\dfrac{2}{7}=\dfrac{10}{7}\)
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\(tan\dfrac{\pi}{6}+tan\dfrac{2\pi}{9}+tan\dfrac{5\pi}{18}+tan\dfrac{\pi}{3}\)\(=\left(tan\dfrac{\pi}{6}+tan\dfrac{\pi}{3}\right)+\left(tan\dfrac{2\pi}{9}+tan\dfrac{5\pi}{18}\right)\) (1)
Áp dụng công thức: \(tanx+tan\left(90^o-x\right)=tanx+cotx=\dfrac{1}{sinx.cosx}\)
Ta được:(1) = \(\dfrac{1}{sin\dfrac{\pi}{6}cos\dfrac{\pi}{6}}+\dfrac{1}{sin\dfrac{2\pi}{9}.cos\dfrac{2\pi}{9}}\)
\(=\dfrac{2}{sin\dfrac{\pi}{3}}+\dfrac{3}{sin\dfrac{4\pi}{9}}\)
Em làm tiếp nhé.