\(sin\alpha=sin\left(180-\alpha\right)=\dfrac{3}{5}\Rightarrow cos\left(180-a\right)=\sqrt{1-sin^2\alpha}=\dfrac{4}{5}\Rightarrow cos\alpha=-\dfrac{4}{5}\)

\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\Rightarrow cot\alpha=-\dfrac{4}{3}\)

\(\Rightarrow A=\dfrac{3.\dfrac{3}{5}-\dfrac{4}{5}}{-\dfrac{3}{4}+\dfrac{4}{3}}=\dfrac{12}{7}\)

\(1.A\cap B=\left(-4;-2\right)\)

\(B/A=\)\([-2;3)\)

\(2.\sqrt{7x^2}-3x=2x\)

\(\Leftrightarrow\sqrt{7x^2}=5x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\7x^2=25x^2\end{matrix}\right.\)\(\Rightarrow x=0\) 

Vậy...

Mất cái đầu vs cuối r bn

Mình lầm vị trí

A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)

A = (-1) \(\times\) 1 \(\times\) (-1)  \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)

A = 1

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{6}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{21}{6}=\dfrac{7}{2}\)

Câu 1:

a) = \(\dfrac{-7}{2}\) x \(\dfrac{45}{32}\) = \(\dfrac{-315}{64}\)

b) = \(\dfrac{18}{7}\) : \(\dfrac{-27}{14}\) = \(\dfrac{18}{7}\) x \(\dfrac{14}{-27}\) = \(\dfrac{-4}{3}\)

c) = \(\dfrac{-3}{8}\) x ( \(\dfrac{5}{11}\) + \(\dfrac{6}{11}\) + 2 ) = \(\dfrac{-3}{8}\) x 3 = \(\dfrac{-9}{8}\)

Câu 2:

\(\dfrac{-3}{5}\) . x + \(\dfrac{7}{6}\) = \(\dfrac{5}{4}\)

\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{5}{4}\) - \(\dfrac{7}{6}\)

\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{1}{12}\)

\(\Leftrightarrow\) x = \(\dfrac{1}{12}\) : \(\dfrac{-3}{5}\)

\(\Leftrightarrow\) x = \(\dfrac{-5}{36}\)