Đầu bài cho v₁ chưa hay bạn viết nhầm là 30km/h ??

Ta có: P-F_A=3,5
    <=> 5-F_A=3,5
    <=> F_A=1,5(N)

Vậy....

a) Gọi độ dài qđ là: s(km), s>0

Ô tô đi nửa qđ đầu mất: \(\dfrac{s}{\dfrac{2}{v_1}}=\dfrac{s}{2v_1}\)(h)

Ô tô đi nửa qđ sau mất: \(\dfrac{s}{\dfrac{2}{v_2}}=\dfrac{s}{2v_2}\)(h)
Vận tốc TB của ng đó trên cả qđ là: \(v_{tb}=\dfrac{s}{\dfrac{s}{2v_1}+\dfrac{s}{2v_2}}=\dfrac{2v_1v_2}{v_1+v_2}\)(km/h)
Vậy......

b) Gọi tổng thời gian ô tô đó chuyển động là t(h), t>0

Quãng đường ô tô đó đi đc trong nửa t.g đầu là: \(\dfrac{t}{2}.v_1\)(km)
Quãng đường ô tô đó đi đc trong nửa t.g sau là: \(\dfrac{t}{2}.v_2\)(km)
Vận tốc TB của ô tô đó là: \(v'_{tb}=\dfrac{\dfrac{t}{2}.v_1+\dfrac{t}{2}.v_2}{t}=\dfrac{v_1+v_2}{2}\)(km/h)
Vậy......

c) Ta có: \(v_{tb}-v'_{tb}=\dfrac{2v_1v_2}{v_1+v_2}-\dfrac{v_1+v_2}{2}=\dfrac{4v_1v_2}{2\left(v_1+v_2\right)}-\dfrac{\left(v_1+v_2\right)^2}{2\left(v_1+v_2\right)}\)

\(=\dfrac{4v_1v_2-\left(v_1+v_2\right)^2}{2\left(v_1+v_2\right)}=\dfrac{-\left(v_1-v_2\right)^2}{2\left(v_1+v_2\right)}\)

Vì \(\left\{{}\begin{matrix}\left(v_1-v_2\right)^2>0\\\left(v_1+v_2\right)>0\left(vì v_1, v_2>0\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\left(v_1-v_2\right)^2< 0\\2\left(v_1+v_2\right)>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{-\left(v_1-v_2\right)^2}{2\left(v_1+v_2\right)}< 0\Rightarrow v_{tb}< v'_{tb}\)

Vậy.....

Gọi số dầu lúc đầu ở thùng A là x(l), x>25
=> số dầu lúc đầu ở thùng b là: 105-x(l)
Ta có: x-25=\(\dfrac{2}{5}\)(105-x+25)
<=> x-25=42-\(\dfrac{2}{5}\)x+10
<=> \(\dfrac{7}{5}\)x=77
<=> x=55(thỏa mãn) 
=>số dầu lúc đầu ở thùng b là 105-55=50(l)

Vậy.....

a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\left(\dfrac{5}{9}\right)^2.\left(\dfrac{4}{7}\right)^2.\left(\dfrac{14}{3}\right)^2}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\left(\dfrac{7}{4}\right)^2.\left(\dfrac{8}{5}\right)^2.\left(\dfrac{14}{9}\right)^2}\)

\(=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{7.3136}{7.81}}=\sqrt{\dfrac{3136}{81}}=\sqrt{\left(\dfrac{56}{9}\right)^2}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810}.\sqrt{\left(11-5\right)\left(11+5\right)}=\sqrt{36.6.81}.\sqrt{6.16}=6.9\sqrt{6}.4\sqrt{6}=1296\)

d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)

\(=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5.1^2=6-2\sqrt{2}+3\sqrt{2}-5\)

\(=1+\sqrt{2}\)

c) \(\left(\dfrac{1}{2}.\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}.\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}.10.\sqrt{2}\right):\dfrac{1}{8}\)

\(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}.\sqrt{2}+8.\sqrt{2}\right):\dfrac{1}{8}=\sqrt{2}\left(\dfrac{1}{4}-\dfrac{3}{2}+8\right).8\)

\(=\sqrt{2}.\dfrac{27}{4}.8=54\sqrt{2}\)

b) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)

a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\left(4-6+2\sqrt{5}\right)-\sqrt{5}=-2+\sqrt{5}\)

d) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)