a)Độ cao thả vật \(h=\dfrac{1}{2}gt^2=\dfrac{1}{2}.10.8^2=320\left(m\right)\)

b) Quãng đường đi được trong 2 giây cuối

S₁ = S_8s - S_6s = \(\dfrac{1}{2}.10.8^2-\dfrac{1}{2}.10.6^2=140\left(m\right)\)

c) Quãng đường rơi trong giây thứ 6

\(S_2=S_{6s}-S_{5s}=\dfrac{1}{2}.10.6^2-\dfrac{1}{2}.10.5^2=55\left(m\right)\)

d) \(v_{5s}=gt=10.5=50\)(m/s)

\(v_{4s}=gt=10.4=40\) (m/s)

\(\Delta v=v_{5s}-v_{4s}=50-40=10\)(m/s)

Xyz#12 ; 5.8 

Xyz#12 ; 5.8 

được có 1 câu làm thế nào 

Đề lỗi rồi 

Mạch : (R_MC//R_NC) nt R₂ nt R₁ 

(V) đo R_DB

C giữa R_MN => \(R_{MC}=R_{NC}=4\Omega\)

\(R_{MC-CN}=\dfrac{R_{MC}.R_{NC}}{R_{MC}+R_{NC}}=\dfrac{4.4}{4+4}=2\Omega\)

\(R_{tđ}=R_1+R_2+R_{MC-CN}=0,4+0,6+2=3\Omega\)

=> \(I_{AB}=\dfrac{U}{R_{tđ}}=\dfrac{24}{3}=8\left(A\right)\)

=> \(U_1=I_{AB}.R_1=0,4.8=3,2\left(V\right)\)

=> \(U_{R_2-Bt}=U_{AB}-U_1=24-3,2=20,8\left(V\right)\)

=> (V) chỉ 20,8 (V) 

ý mình chỗ mũi tên gạch gần điểm B là bạn vẽ sai hay là vẽ chiều dòng điện 

Ta có \(A\sqrt{\dfrac{17}{4}}=\sqrt{\dfrac{17}{4}\left(a^2+\dfrac{1}{b^2}\right)}+\sqrt{\dfrac{17}{4}\left(b^2+\dfrac{1}{c^2}\right)}+\sqrt{\dfrac{17}{4}\left(c^2+\dfrac{1}{a^2}\right)}\)

Có :\(\sqrt{\dfrac{17}{4}\left(a^2+\dfrac{1}{b^2}\right)}=\sqrt{\left[2^2+\left(\dfrac{1}{2}\right)^2\right]\left[\left(\dfrac{1}{b}\right)^2+a^2\right]}\ge\sqrt{\left(\dfrac{2}{b}+\dfrac{a}{2}\right)^2}\)

= \(\dfrac{2}{b}+\dfrac{a}{2}\)

Tương tự đươc \(\dfrac{A\sqrt{17}}{2}\ge\dfrac{2}{b}+\dfrac{a}{2}+\dfrac{2}{c}+\dfrac{b}{2}+\dfrac{2}{a}+\dfrac{c}{2}\)

\(=\left(\dfrac{a}{2}+\dfrac{1}{8a}\right)+\left(\dfrac{b}{2}+\dfrac{1}{8b}\right)+\left(\dfrac{c}{2}+\dfrac{1}{8c}\right)+\dfrac{15}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\ge2\sqrt{\dfrac{a}{2}.\dfrac{1}{8a}}+2\sqrt{\dfrac{b}{2}.\dfrac{1}{8b}}+2\sqrt{\dfrac{c}{2}.\dfrac{1}{8c}}+\dfrac{15}{8}.\dfrac{9}{a+b+c}\)

\(\ge\dfrac{3}{2}+\dfrac{15}{8}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{51}{4}\Leftrightarrow A\ge\dfrac{3\sqrt{17}}{2}\)

"=" KHI a = b = c = 1/2

vẽ lại mạch ik bạn 

đoạn cuối có điểm B mk không hiểu 

\(P=\sqrt{\left(a+b\right)^2+b^2}+\sqrt{\left(b+c\right)^2+c^2}+\sqrt{\left(a+c\right)^2+a^2}\)

\(\Leftrightarrow\sqrt{5}P=\sqrt{5\left[\left(a+b\right)^2+b^2\right]}+\sqrt{5\left[\left(b+c\right)^2+c^2\right]}+\sqrt{5\left[\left(a+c\right)^2+a^2\right]}\)

Có : \(\sqrt{5\left[\left(a+b\right)^2+b^2\right]}=\sqrt{\left[\left(a+b\right)^2+b^2\right]\left(2^2+1^2\right)}\ge2\left(a+b\right)+b\)

(BĐT Bunyakovski) 

Tương tự được \(\sqrt{5}P\ge2\left(a+b\right)+b+2\left(b+c\right)+c+2\left(a+c\right)+a\)

\(\Leftrightarrow\sqrt{5}P\ge5\left(a+b+c\right)\Leftrightarrow P\ge\sqrt{5}\)

"=" khi \(\left\{{}\begin{matrix}\dfrac{a+b}{2}=b\\\dfrac{b+c}{2}=c\\\dfrac{a+c}{2}=a\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)