1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

em tán thành 

vì Kỉ luật tạo nên sự đoàn kết, thống nhất, đảm bảo dân chủ được thực hiện một cách tốt nhất.

Có tinh thần, trách nhiệm trong học tập và công việc, luôn có ý thức vì tập thể.

Câu 4: 

a: Xét ΔABD và ΔAED có 

AB=AE

\(\widehat{BAD}=\widehat{EAD}\)

AD chung

Do đó: ΔABD=ΔAED

Câu 1:

\(a,=\dfrac{1}{2}+9\cdot\dfrac{1}{9}-18=\dfrac{1}{2}+1-18=-\dfrac{33}{2}\\ b,=2-1+4\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{9}\cdot9=1+1+9=11\\ c,=-21,3\left(54,6+45,4\right)=-21,3\cdot100=-2130\\ d,B=\left(\dfrac{1}{16}+\dfrac{1}{2}-\dfrac{1}{16}\right):\left(\dfrac{1}{8}-\dfrac{1}{8}+1\right)=\dfrac{1}{2}:1=\dfrac{1}{2}\)

Độ dãn của lò xo:

\(F_{đh}=k\cdot\Delta l\Rightarrow\Delta l=\dfrac{F_{đh}}{k}=\dfrac{2}{100}=0,02\)m=2cm

B ... Thì lo xo dài 35cm tính f2

b)Lò xo dài 35cm.

   \(\Rightarrow\) Độ dãn của lò xo: \(\Delta l'=l'-l_0=35-30=5cm=0,05m\)

   \(F_2=k\cdot\Delta l'=100\cdot0,05=5N\)

Bài 9:

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{z}{-17}=\dfrac{t}{9}=-2\)

=>x=-10; y=6; z=34; t=-18

Bài 10:

\(\Leftrightarrow\dfrac{8}{x}=\dfrac{y}{21}=\dfrac{40}{z}=\dfrac{16}{t}=\dfrac{u}{111}=\dfrac{4}{3}\)

=>x=6; y=28; z=30; t=12; u=148

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)

a) \(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(A=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(A=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(A=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(A=\dfrac{1-3}{1+5}\)

\(A=-\dfrac{2}{6}\)

\(A=-\dfrac{1}{3}\)

b) \(B=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(B=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\cdot\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(B=\dfrac{1}{4}+\dfrac{3\cdot\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\cdot\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(B=\dfrac{1}{4}+\dfrac{3}{4}\)

\(B=\dfrac{1+3}{4}\)

\(B=\dfrac{4}{4}\)

\(B=1\)