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Em đăng kí nhận GP

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\(14-8x^2=\dfrac{3}{2}\\ =>8x^2=14-\dfrac{3}{2}\\ =>8x^2=\dfrac{25}{2}\\ =>x^2=\dfrac{25}{2}:8\\ =>x^2=\dfrac{25}{16}\\ =>x=\pm\dfrac{5}{4}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow1+2^2=\dfrac{1}{cos^2\alpha}\\ \Rightarrow5=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{5}\)

Bạn xem lại đề bài nhé.

\(\dfrac{3}{2}-\left(x-2\dfrac{1}{3}\right)=\dfrac{-3}{4}-1\dfrac{1}{8}\\ \Rightarrow\dfrac{3}{2}-x+\dfrac{7}{3}=\dfrac{-3}{4}-\dfrac{9}{8}\\ \Rightarrow\dfrac{23}{6}-x=-\dfrac{15}{8}\\ \Rightarrow x=\dfrac{23}{6}-\left(-\dfrac{15}{8}\right)\\ \Rightarrow x=\dfrac{137}{24}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\\ \Rightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9\dfrac{5}{7}-\dfrac{5}{7}\\ \Rightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9\\ \Rightarrow x-\dfrac{1}{2}=9.\dfrac{1}{3}\\ \Rightarrow x-\dfrac{1}{2}=3\\ \Rightarrow x=3+\dfrac{1}{2}\\ \Rightarrow x=\dfrac{7}{2}\)

\(a/5.3^{x+3}-14=121\\ \Rightarrow5.3^{x+3}=135\\ \Rightarrow3^{x+3}=135:5=27\\ \Rightarrow3^{x+3}=3^3\\ \Rightarrow x+3=3\\ \Rightarrow x=0\)

\(b/4^x=4.4^{10}-3.4^{10}\\ \Rightarrow4^x=\left(4-3\right).4^{10}\\ \Rightarrow4^x=4^{10}\\ \Rightarrow x=10\)

\(\left(x+5\right)^3.\left(2x\right)^3=\left[2x\left(x+5\right)\right]^3\\ =\left(2x^2+10x\right)^3\)