Đo đi em?!?!

1cm

mù chữ hả em:)?? Lấy thước ra mà tự đo:/

\(A=log_m\left(8m\right)=log_mm+log_m8\)

\(=1+log_m8\)

\(=1+\dfrac{1}{log_8m}=1+\dfrac{1}{log_{2^3}m}=1+\dfrac{1}{\dfrac{1}{3}\cdot log_2m}\)

\(=1+\dfrac{1}{\dfrac{1}{3}a}=1+1:\dfrac{a}{3}=1+\dfrac{3}{a}=\dfrac{a+3}{a}\)

=>Chọn A

Bạn ghi đề mới biết chứ

12.(x-1)=64+8

12.(x-1)=72

     x-1=72:12

     x-1=6

         x=6+1

         x=7

học tốt...

12 . ( x - 1 ) : 3 = 4³ + 2³

12 . ( x - 1 ) : 3 = 64 + 8

12 . ( x - 1 ) : 3 = 72

12 . ( x - 1 ) = 72 . 3

12 . ( x - 1 ) = 216

x - 1 = 216 : 12

x - 1 = 18

x = 18 + 1

x = 19

=19

hok tốt

Hai tam giác vuông CAB và CFE đồng dạng (chung góc C)

\(\Rightarrow\dfrac{CF}{CA}=\dfrac{EF}{AB}=\dfrac{AD}{AB}=\dfrac{AD}{3}\)

\(\Rightarrow\dfrac{AC-AF}{AC}=\dfrac{AD}{3}\Leftrightarrow\dfrac{AC-2}{AC}=\dfrac{AD}{3}\Rightarrow AD=3\left(\dfrac{AC-2}{AC}\right)\)

\(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{3}{2}AC\)

\(S_{ADEF}=AD.AF=2AD=6\left(\dfrac{AC-2}{AC}\right)\)

Theo đề bài: \(S_{ADEF}=\dfrac{1}{2}S_{ABC}\Rightarrow6\left(\dfrac{AC-2}{AC}\right)=\dfrac{1}{2}.\dfrac{3}{2}AC\)

\(\Leftrightarrow8\left(AC-2\right)=AC^2\Leftrightarrow AC^2-8AC+16=0\)

\(\Leftrightarrow\left(AC-4\right)^2=0\Leftrightarrow AC=4\)

Vậy \(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.3.4=6\left(cm^2\right)\) \(\Rightarrow S_{ADEF}=3\)

Chọn A

1. a

\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)

\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)

\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)

1.b

\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)

\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)

\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)

2.a

\(\dfrac{4}{5}+x=\dfrac{11}{10}\)

\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)

2.b

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

\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)

=80-[130-8^2]

^=80-[130-64]

^{=80 - 66}

⁼¹⁴

d) 80 – [130 – (12 – 4)²] = 80 - (130 - 64) = 80 - 66 = 14.

tích mik na

d) 80 – [130 – (12 – 4)²] = 80 - (130 - 64) = 80 - 66 = 14.