a) 15 + 4.12 - 3²

=15 + 48 - 9

= 63 - 9

= 54

b) {[(37 + 13) : 5].2 - 45 : 5}.7 - 2022⁰

= [(50 : 5).2 - 9].7 - 1

= (10.2 - 9).7 - 1

= (20 - 9).7 - 1

= 11.7 - 1

= 77 - 1

= 76

\(15+4.12-3^2=15+48-9=63-9=54\\ ---\\ \left\{\left[\left(37+13\right):5\right]-45:5\right\}.7-2022^0=\left\{\left[50:5\right]-9\right\}.7-1\\ =\left\{10-9\right\}.7-1\\ =1.7-1=6\)

bài 1: 

\(\left\{{}\begin{matrix}x+y=57\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=228\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6y=234\\x+y=57\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=39\\x=18\end{matrix}\right.\)

a, 98; 56; 24

    98 = 2.7²

    56 = 2³.7

    24 = 2³.3

BCNN(98; 56;24) = 2³.3.7² = 1176

b, 50; 600; 120

50 = 2.5²

600 = 2³.3.5²

120 = 2³.3.5

BCNN(50; 600;120) =2³.3.5²= 600 

c, 168; 120; 144

168 = 2³.3.7

120 = 2³.3.5

144 = 2⁴.3²

BCNN(168; 120; 144) = 2⁴.3².5.7 = 5040

A = 3\(x^2\) + 9\(x\) - 7

A = 3.(\(x^2\) + 3\(x\) + \(\dfrac{9}{4}\)) - 7

A = 3.(\(x\) + \(\dfrac{3}{2}\))² - \(\dfrac{55}{4}\)

Vì (\(x\) + \(\dfrac{3}{2}\))² ≥ 0; ⇒ 3.(\(x\) + \(\dfrac{3}{2}\))² - \(\dfrac{55}{4}\) ≥ - \(\dfrac{55}{4}\)

A(min) = - \(\dfrac{55}{4}\) ⇔ \(x\) + \(\dfrac{3}{2}\) = 0 ⇔ \(x\) = - \(\dfrac{3}{2}\) 

1. will be having

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a) \(\left(0,5\right)^{12}:\left(0,5\right)^{10}=\left(0,5\right)^{12-10}=\left(0,5\right)^2\)

b) \(\sqrt{36}=\pm6\)

c)\(\left(0,75\right)^{22}:\left(0,75\right)^{12}=\left(0,75\right)^{22-12}=\left(0,75\right)^{10}\)

d) \(\sqrt{49}=\pm7\)

a.

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sin4x+\dfrac{\sqrt{2}}{2}cos4x=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos4x.cos\left(\dfrac{\pi}{4}\right)+sin4x.sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{4}=arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\\4x-\dfrac{\pi}{4}=-arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{16}-\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{3}\right)+sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x-\dfrac{\pi}{3}=-arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x=\dfrac{\pi}{3}-arrcos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{6}\right)+sinx.sin\left(\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{3\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

\(\left(2-\dfrac{1}{2}\right)\cdot\left(\dfrac{-3}{4}+\dfrac{1}{2}\right)\)

\(=\left(\dfrac{4}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{-3}{4}+\dfrac{2}{4}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{-1}{4}\)

\(=\dfrac{-3}{8}\)