Xét \(2\Delta:\Delta ABE\) và \(\Delta ACO\) có:
\(\left\{{}\begin{matrix}\widehat{A}chung\\\widehat{AEB}=\widehat{AOC}\end{matrix}\right.\)

\(\Rightarrow\Delta ABE\sim\Delta ACO\)

\(\Rightarrow\dfrac{AB}{AC}=\dfrac{EB}{OC}\Leftrightarrow\dfrac{AB}{AB+BC}=\dfrac{EB}{OC}\Leftrightarrow\dfrac{AB}{AB+125}=\dfrac{20}{25}\)

\(\Leftrightarrow AB=500\left(m\right)\)

\(V=\pi\int\limits^9_4\left(\sqrt{2x-1}\right)^2dx=\pi\left[\left(9^2-9\right)-\left(4^2-4\right)\right]=60\pi\)

Ta có: \(v\left(t\right)=s'\left(t\right)=t^2-3t+10\)

Vật đạt vận tốc 20m/s trong thời gian:

\(t^2-3t+10=20\)

\(\Leftrightarrow t^2-3t-10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Thay x = 5 vào s(t), ta được: \(\dfrac{1}{3}.5^3-\dfrac{3}{2}.5^2+10.5+2\approx56,2\)

Vậy ...

Ta có: \(\left\{{}\begin{matrix}BM=MC\\MF//AB\left(F\in AC\right)\end{matrix}\right.\)

=> AF = FC

Lại có G là giao điểm AM và BF

=> G là trọng tâm tam giác ABC

\(\Rightarrow AG=\dfrac{2}{3}AM\)

Lại có: \(AM=\dfrac{1}{2}BC\)

\(\Rightarrow BC=3AG\)

\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^1=\dfrac{3}{8}\)

\(\Leftrightarrow2.\left(\dfrac{1}{2}\right)^x=-\dfrac{1}{8}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^x=-\dfrac{1}{16}\)

=> Vô nghiệm

pthh: \(Fe+2HCl--->FeCl_2+H_2\uparrow\)

a. \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b. \(n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c. \(n_{HCl}=2.0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

1.B

2.B

3.C

4.A

5.C

6.D

7.C

8.D

9.D

10.C

\(A=\left|x-7\right|+2x-3=x-7+2x-3=3x-10\)