Gọi vận tốc lúc đi là \(x\left(km/h\right)\left(x>0\right)\)

Vận tốc lúc về là \(x+6\left(km/h\right)\)

Thời gian đi là \(\dfrac{35}{x}\left(h\right)\)

Thời gian về là \(\dfrac{42}{x+6}\left(h\right)\)

Vì thời gian về bằng \(\dfrac{12}{13}\) thời gian đi

nên ta có pt:

\(\dfrac{12}{13}\cdot\dfrac{35}{x}=\dfrac{42}{x+6}\\ \Leftrightarrow\dfrac{420}{13x}=\dfrac{42}{x+6}\\ \Leftrightarrow13x=10\left(x+6\right)\\ \Leftrightarrow13x=10x+60\\ \Leftrightarrow3x=60\\ \Leftrightarrow x=20\left(T/m\right)\)

Vậy vận tốc lúc đi là \(20\left(km/h\right)\)

Vận tốc lúc đi là \(20+6=26\left(km/h\right)\)

câu d giống câu a nha

\(C=\sqrt{x-2}+\sqrt{y-3}\\ \Rightarrow C^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\\ =x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\\ =\left(x+y\right)-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\\ =6-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\\ =1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

Áp dụng \(BDT:Cô-si\)

\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\le1+\left(x-2+y-3\right)\\ =1+\left(x+y-5\right)\\ =1+\left(6-5\right) =2\\ \Rightarrow C^2\ge2\\ \Rightarrow C\ge\sqrt{2}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6+\left(-1\right)}{2}=\dfrac{5}{2}\\y=\dfrac{6-\left(-1\right)}{2}=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(C_{Max}=\sqrt{2}\) khi \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)

11 đoạn thẳng

bằng 3,6x3,6

2 cách đó ko saj đâu nha

\(P=a^3+\dfrac{1}{b^3}\\ =\left(a+\dfrac{1}{b}\right)\left(a^2-ab+\dfrac{1}{b^2}\right)\\ =\left(a+\dfrac{1}{b}\right)\left(a^2+2ab+\dfrac{1}{b^2}-3ab\right)\\ =\left(a+\dfrac{1}{b}\right)\left[\left(a+\dfrac{1}{b}\right)^2-3ab\right]\\ =-4\left[\left(-4\right)^2-3\left(-4\right)\right]=-112\)

dap an la : 1011;1012;1013;1014;1015 ..............

  ung ho minh nhe 

\(P=\dfrac{x^2+x+1}{x^2+2x+1}\\ =\dfrac{x^2+2x-x+1-1+1}{x^2+2x+1}\\ =\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\\ =1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\)

Đặt \(\dfrac{1}{x+1}=t\)

\(\Rightarrow P=t^2-t+1\\ =t^2-t+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(t^2-t+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Do \(\left(t-\dfrac{1}{2}\right)^2\ge0\forall x;t\)

\(\Rightarrow P=\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;t\)

Dấu "=" xảy ra khi:

\(\left(t-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow t-\dfrac{1}{2}=0\\ \Leftrightarrow t=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}\\ \Leftrightarrow x+1=2\\ \Leftrightarrow x=1\)

Vậy \(P_{Min}=\dfrac{3}{4}\) khi \(x=1\)