1đvc = \(0,16605.10^{-23}\)

\(M_B=\dfrac{2,6568}{0,16605}=16\Rightarrow B\) là O

\(M_A=32\Rightarrow\) A là S 

56.a

57.c

58.a

59.d

60.b

61.d

62.c

63.a

64.c

65.c

Tuổi bố 3 năm sau : \(\dfrac{46+28}{2}=37\) ( tuổi )

Tuổi bố hiện nay : 37 - 3 = 34 ( tuổi )

Tuổi con hiện nay : 34 - 28 = 6 ( tuổi ) 

Đ/s: ... 

Nhanh thế 

v < 0 vì đồ thị đi xuống dưới 

a. ... \(\Leftrightarrow x+5=\dfrac{2}{3}\Leftrightarrow x=-\dfrac{13}{3}\)

b. ... \(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{5}{3}\\x-2=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

c. ... \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d. ... \(\Leftrightarrow2x-1=-\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)

e. ... \(\Leftrightarrow3x-2=-4\Leftrightarrow x=-\dfrac{2}{3}\)

g. ... \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{5}=\dfrac{7}{5}\\\dfrac{1}{3}x+\dfrac{1}{5}=-\dfrac{7}{5}\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{18}{5}\\x=-\dfrac{24}{5}\end{matrix}\right.\)

Số học sinh không tham gia tiết mục nào : \(50-20-17+8=21\) ( học sinh )

Đ/s : ... 

a. \(\left(-\dfrac{2}{3}\right)^4=\dfrac{2^4}{3^4}=\dfrac{16}{81}\)

b. \(\left(-3\dfrac{1}{3}\right)^2=\left(-\dfrac{10}{3}\right)^2=\dfrac{100}{9}\)

c. \(\left(-1\dfrac{5}{7}\right)^2=\left(-\dfrac{12}{7}\right)^2=\dfrac{144}{49}\)

d. \(\left(-1\dfrac{3}{4}\right)^3=\left(-\dfrac{7}{4}\right)^3=-\dfrac{343}{64}\)

e. \(\left(-0,4\right)^2=0,16\)

\(2^x=16\Rightarrow x=4\)

\(3^{x+1}=9^x\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)

\(2^{x-1}=8^5\Rightarrow2^{x-1}=2^{15}\Rightarrow x-1=15\Rightarrow x=16\)

\(\left\{{}\begin{matrix}7x=3y\\y-x=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{7x}{3}\\\dfrac{7x}{3}-x=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=28\\x=12\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2x}{5y}=-\dfrac{1}{6}\\2x-5y=14\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5y=-12x\\2x+12x=14\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}y=-\dfrac{12}{5}\\x=1\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{7}=\dfrac{x+y}{7}=\dfrac{14}{7}=2\Rightarrow x=6;y=8;z=14\)

a. \(\sqrt{13}+3>\sqrt{11}+2\sqrt{2}\)  ( 3 > 2\(\sqrt{2}\) )

b. \(\sqrt{13}-\sqrt{11}-\left(\sqrt{7}-\sqrt{5}\right)\) = \(\dfrac{2}{\sqrt{13}+\sqrt{11}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}< 0\)

\(\Rightarrow\sqrt{13}+\sqrt{5}< \sqrt{11}+\sqrt{7}\)