a/ \(s=\dfrac{1}{2}at^2=\dfrac{1}{2}.2.3^2=9\left(m\right)\)

b/ \(A=P.h=50.9=450\left(N\right)\)

c/ \(F=am=5.2=10\left(N\right)\)

\(A=F.s=10.9=90\left(N\right)\)

Ta có :

\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)

\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)

Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)

Vậy...

\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)

Ta có :

+) \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}\)

+) \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}\)

\(\Leftrightarrow S< \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{40}+\dfrac{1}{40}\)

\(\Leftrightarrow S< \dfrac{1}{2}\)

Vậy,,,

Ta có :

\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)

\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)

\(\Leftrightarrow A>B\)

Ta có :

+) \(-\dfrac{12}{n}\in Z\Leftrightarrow n\inƯ\left(-12\right)\) \(\Leftrightarrow n\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\) \(\left(1\right)\)

+) \(\dfrac{15}{n-2}\in Z\Leftrightarrow n-2\inƯ\left(15\right)\) \(\Leftrightarrow n-2\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

\(\Leftrightarrow n\in\left\{-13;-3;-1;1;3;5;7;17\right\}\) \(\left(2\right)\)

+) \(\dfrac{8}{n+1}\in Z\) \(\Leftrightarrow n+1\inƯ\left(8\right)\) \(\Leftrightarrow n+1\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n\in\left\{-9;-5;-3;-2;0;1;3;7\right\}\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\) \(\Leftrightarrow n\in\left\{1;3\right\}\)

Vậy...

Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\) (tự cm)

Lại có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)

Áp dụng BĐT trên ta có : : \(xy\le\left(\dfrac{x+y}{2}\right)^2\)

\(\Leftrightarrow A\ge\dfrac{x+y}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{1}{\dfrac{1}{2^2}}=4\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy...

a/ \(v^2-v_o^2=2as\) \(\Leftrightarrow a=\dfrac{1,2^2}{2.5}=0,576\left(m\backslash s^2\right)\)

\(s=\dfrac{1}{2}at^2\Leftrightarrow t=\dfrac{25}{6}\left(s\right)\)

\(P=\dfrac{F.s}{t}=\dfrac{a.m.s}{t}=\dfrac{0,576.10.5}{\dfrac{25}{6}}=6,972\left(W\right)\)

\(\left(x^2-2x-3\right)^2\ge\left(x^2+3x+3\right)^2\)

\(\Leftrightarrow\left(x^2-2x-3\right)^2-\left(x^2+3x+3\right)^2\ge0\)

\(\Leftrightarrow\left(-5x-6\right)\left(2x^2+x\right)\ge0\)

\(\Leftrightarrow x\left(-5x-6\right)\left(2x+1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{6}{5}\le x\le0\\\left(-\infty;-\dfrac{1}{2}\right)\end{matrix}\right.\)

a/ Thời gian đi hết đoạn đường AB :

\(t=\dfrac{s}{v}=\dfrac{45}{30}=1,5\left(h\right)\)

b/ Để đến AB sớm hơn 30p ô tô phải đi với thời gian là \(1h\)

Vận tốc ô tô cần đi khi đó :

\(v'=\dfrac{s}{t'}=\dfrac{45}{1}=45\left(km\backslash h\right)\)

a/ \(v^2-v_o^2=2as\)

\(\Leftrightarrow10^2-15^2=2a.125\)

\(\Leftrightarrow a=\dfrac{10^2-15^2}{2.125}=-0,5\left(m\backslash s^2\right)\)

Thời gian kể từ lúc tắt máy đến khi dừng lại :

\(v=v_o+at\) \(\Leftrightarrow t=\dfrac{v-v_o}{a}=\dfrac{0-15}{-0,5}=30\left(s\right)\)

b/ Quãng đường từ lúc tắt máy đến khi dừng lại :

\(s=\dfrac{v^2-v_o^2}{2a}=\dfrac{0-15^2}{2.\left(-0,5\right)}=225\left(m\right)\)