k: \(\left(4x-16\right)\left(-72+9x\right)=0\)

=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)

=>\(36\left(x-4\right)\left(x-8\right)=0\)

=>\(\left(x-4\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)

m: \(\left(20+5x\right)\left(4x-8\right)=0\)

=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)

=>\(\left(x+4\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

n: \(\left(-4x+48\right)\left(2x-24\right)=0\)

=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)

=>\(\left(x-12\right)^2=0\)

=>x-12=0

=>x=12

o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)

=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)

p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)

=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)

=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)

q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)

=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)

=>\(x\left(x-2\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)

r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)

=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)

=>\(x\left(x+3\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)

s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)

=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)

=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)

t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)

=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)

=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)

x + 3y = x(5y - 1)   (1)

1/x - 3/y = -2    (2)

(1) ⇔ x(5y - 1) - x = 3y

⇔ x(5y - 2) = 3y

⇔ x = 3y/(5y - 2)     (3)

Thế (3) vào (2) ta được:

(2) ⇔ 1/[3y/(5y - 2)] - 3/y = -2

⇔ (5y - 2)/3y - 3/y = -2

⇔ 5y - 2 - 9 = -6y

⇔ 5y + 6y = 11

⇔ 11y = 11

⇔ y = 1 thế vào (3) ta được:

x = 3.1/(5.1 - 2) = 1

Vậy S = {(1; 1)}

8:

\(=\dfrac{cos10-\sqrt{3}\cdot sin10}{sin10\cdot cos10}=\dfrac{2\left(\dfrac{1}{2}\cdot cos10-\dfrac{\sqrt{3}}{2}\cdot sin10\right)}{sin20}=\dfrac{sin\left(30-10\right)}{sin20}=1\)

10:

\(=\left(2-\sqrt{3}\right)^2+\left(2+\sqrt{3}\right)^2\)

=7-4căn 3+7+4căn 3=14

12:

\(=cos^270^0+\dfrac{1}{2}\left[cos60-cos140\right]\)

\(=cos^270^0+\dfrac{1}{2}\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot2cos^270^0+\dfrac{1}{.2}\)

=1/4+1/2=3/4

Bài 1:

a, \(\dfrac{2}{3}\) + \(\dfrac{1}{5}\). \(\dfrac{10}{7}\)

= \(\dfrac{2}{3}\) + \(\dfrac{2}{7}\) 

= \(\dfrac{20}{21}\)

b, \(\dfrac{7}{12}\) - \(\dfrac{27}{7}\). \(\dfrac{1}{18}\)

= \(\dfrac{7}{12}\) - \(\dfrac{3}{14}\)

= \(\dfrac{31}{84}\)

c, \(\dfrac{3}{10}\). \(\dfrac{-5}{6}\) - \(\dfrac{1}{8}\)

= - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\)

= - \(\dfrac{3}{8}\)

d, - \(\dfrac{4}{9}\): \(\dfrac{8}{3}\) + \(\dfrac{1}{18}\)

= - \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\)

= - \(\dfrac{1}{9}\)

e,  {[(\(\dfrac{1}{2}\) - \(\dfrac{2}{3}\))² : 2 ] - 1}. \(\dfrac{4}{5}\)

= {[ (-\(\dfrac{1}{6}\))² : 2] - 1}. \(\dfrac{4}{5}\)

= { [\(\dfrac{1}{36}\) : 2] - 1}. \(\dfrac{4}{5}\)

= { \(\dfrac{1}{72}\) - 1}. \(\dfrac{4}{5}\)

=- \(\dfrac{71}{72}\).\(\dfrac{4}{5}\)

= -\(\dfrac{71}{90}\)

g, [(\(\dfrac{32}{25}\) +1): \(\dfrac{2}{3}\)].(\(\dfrac{3}{4}\) - \(\dfrac{1}{8}\))²

= [ \(\dfrac{57}{25}\) : \(\dfrac{2}{3}\)].(\(\dfrac{5}{8}\))²

= \(\dfrac{171}{50}\). \(\dfrac{25}{64}\)

= \(\dfrac{171}{28}\)

Mọi người làm lời giải chi tiết ra giúp em với ạ

\(A=-2\left(x^2-\dfrac{1}{2}x\right)=-2\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=-2\left(x^2-2x.\dfrac{1}{4}+\dfrac{1}{16}\right)+\dfrac{1}{8}=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{8}\le\dfrac{1}{8}\)

\(\Rightarrow A_{max}=\dfrac{1}{8}\)

a) \(D=4\sqrt{\dfrac{1}{3}}+5\sqrt{12}-6\sqrt{27}\)

\(=\dfrac{4}{9}\sqrt{3}+5.2\sqrt{3}-6.3\sqrt{3}\)

\(=\dfrac{4}{9}\sqrt{3}+10\sqrt{3}-18\sqrt{3}\)

\(=-\dfrac{68}{9}\sqrt{3}\)

b) \(E=\dfrac{2}{\sqrt{3}-1}-\sqrt{4-2\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}\)

\(=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)\)

\(=\sqrt{3}+1-\sqrt{3}+1=2\)

c) \(F=\dfrac{\sqrt{15}-\sqrt{10}}{\sqrt{3}-\sqrt{2}}+\dfrac{3}{2-\sqrt{5}}\)

 \(=\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}+\dfrac{3\left(2+\sqrt{5}\right)}{-1}\)

\(=\sqrt{5}-6-3\sqrt{5}=-2\sqrt{5}-6\)