Bài làm :

Gọi thời gian đi hết cả đoạn đường là 2t

Ta có :

S = v₁t + v₂t = t(v₁ + v₂ )

Vậy vận tốc trung bình trên cả quãng đường là :

\(V_{TB}=\frac{S}{2t}=\frac{t\left(v_1+v_2\right)}{2t}=\frac{v_1+v_2}{2}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ta có :

\(t1=\frac{S1}{v1}=\frac{S}{2v1}\)

\(t2=\frac{S2}{v2}=\frac{S}{2v2}\)

Vận tốc trung bình của người đó là :

\(Vtb=\frac{S}{t1+t2}\times\frac{S}{\frac{S}{2v1}+\frac{S}{2v2}}=\frac{1}{\frac{1}{2v1}+\frac{1}{2v2}}=\frac{1}{\frac{v2+v1}{2v1v2}}\)

a: \(\frac{-2x+3}{5xy}+\frac{3x-4}{5xy}\)

\(=\frac{-2x+3+3x-4}{5xy}=\frac{x-1}{5xy}\)

b: \(\frac{x\left(x-2\right)}{x-3}+\frac{3}{3-x}\)

\(=\frac{x^2-2x-3}{x-3}\)

\(=\frac{\left(x-3\right)\left(x+1\right)}{x-3}=x+1\)

c: \(\frac{2x+7}{3\left(x+2\right)}-\frac{x+5}{3\left(x+2\right)}\)

\(=\frac{2x+7-x-5}{3\left(x+2\right)}\)

\(=\frac{x+2}{3\left(x+2\right)}=\frac13\)

d: \(\frac{5}{2x+6}+\frac{3}{x^2-9}\)

\(=\frac{5}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5\left(x-3\right)+6}{2\left(x-3\right)\left(x+3\right)}=\frac{5x-9}{2\left(x-3\right)\left(x+3\right)}\)

e: \(\frac{3x}{x+2}+\frac{x+3}{x^2-4}\)

\(=\frac{3x\left(x-2\right)+x+3}{\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x+x+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x^2-5x+3}{\left(x-2\right)\left(x+2\right)}\)

f: \(\frac{3x+2}{\left(x+2\right)^2}-\frac{1}{x+2}\)

\(=\frac{3x+2-\left(x+2\right)}{\left(x+2\right)^2}=\frac{2x}{\left(x+2\right)^2}\)

=10000 nha 

jk mk nha 

\(73^2+27^2+54\cdot73=73^2+2\cdot73\cdot27+27^2\)

                                      \(=\left(73+27\right)^2=100^2=10000\)

\(AC=\sqrt{5^2+12^2}=13\left(cm\right)\)

Sxq=(5+12+13)*8=8*30=240cm²

Stp=240+2*12*5=360cm²

Bài 2:

a: \(f\left(x\right)=x^3-9x^2+27x-27\)

\(=x^3-3x^2-6x^2+18x+9x-27\)

\(=x^2\left(x-3\right)-6x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\)

b: \(g\left(x\right)=x^2-6x+9\)

\(=x^2-2\cdot x\cdot3+3^2\)

\(=\left(x-3\right)^2\)

=>f(x)⋮g(x)

=>f(x) chia g(x) thì dư 0

\(\frac{f\left(x\right)}{k\left(x\right)}=\frac{x^3-9x^2+27x-27}{x^2-6x+10}\)

\(=\frac{x^3-6x^2+10x-3x^2+18x-30-x+3}{x^2-6x+10}=x-3+\frac{-x+3}{x^2-6x+10}\)

=>f(x) chia k(x) dư -x+3

\(x\left(x+2\right)-\left(x-2\right)^2=8\)

\(x^2+2x-x^2+4x-4=8\)

\(6x-4=8\)

\(x=2\)