\(3^x+3^{x+2}+3^{x+4}=2457\)

\(\Leftrightarrow3^x+3^2.3^x+3^4.3^x=2457\)

\(\Leftrightarrow3^x.\left(1+3^2+3^4\right)=2457\)

\(\Leftrightarrow3^x.91=2457\)

\(\Leftrightarrow3^x=2457:91\)

\(\Leftrightarrow3^x=27=3^3\)

\(\Leftrightarrow x=3\)

\(3^x+3^{x+2}+3^{x+4}=2457\)

\(\Leftrightarrow3^x+3^x.9+3^x.81=2457\)

\(\Leftrightarrow3^x\left(1+9+81\right)=2457\)

\(\Leftrightarrow3^x.91=2457\)

\(\Leftrightarrow3^x=27\Leftrightarrow3^x=3^3\Leftrightarrow x=3\)

\(\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{13}{15\cdot4}\\ =\dfrac{5}{2}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{13}{60}\\ =\dfrac{193}{60}\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

 \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

......

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )

Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

         \(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)

         \(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)

  .........................

         \(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)

=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

=> D <  1 - \(\dfrac{1}{10}\)

=>D < \(\dfrac{9}{10}\)

=> D < \(\dfrac{10}{10}\)

 Vậy D < 1

\(A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1\)

\(A=458+\dfrac{458}{2}+....+\dfrac{458}{456}+\dfrac{458}{457}-\dfrac{458}{458}\)

\(A=458\left(\dfrac{1}{2}+...+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\right)\)

Ta xét \(\dfrac{1}{2}+....+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\)có :

\(\dfrac{1}{2}=\dfrac{1}{2}\)

\(\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{1}{2}\)

\(\dfrac{1}{9}+\dfrac{1}{10}+....+\dfrac{1}{16}>\dfrac{1}{16}+....+\dfrac{1}{16}=\dfrac{1}{2}\)

\(\dfrac{1}{17}+\dfrac{1}{18}+....+\dfrac{1}{32}>\dfrac{1}{32}+.....+\dfrac{1}{32}=\dfrac{1}{2}\)

\(\dfrac{1}{33}+\dfrac{1}{34}+....+\dfrac{1}{64}>\dfrac{1}{64}+....+\dfrac{1}{64}=\dfrac{1}{2}\)

\(\dfrac{1}{65}+\dfrac{1}{66}+.....+\dfrac{1}{128}>\dfrac{1}{128}+....+\dfrac{1}{128}=\dfrac{1}{2}\)

\(\dfrac{1}{129}+\dfrac{1}{130}+.....+\dfrac{1}{256}>\dfrac{1}{256}+....+\dfrac{1}{256}=\dfrac{1}{2}\)

\(\dfrac{1}{257}+\dfrac{1}{258}+....+\dfrac{1}{458}>\dfrac{1}{458}+...+\dfrac{1}{458}=\dfrac{1}{2}\)

Vậy ta thấy được rằng

\(\dfrac{1}{2}+...+\dfrac{1}{456}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{202}{458}\)

\(=4+\dfrac{202}{458}=\dfrac{2034}{458}\)

Vậy \(A>458.\dfrac{2034}{458}=2034\)

Hay tức là A > 2016 ( đpcm )

A = 2(2x + 3)² + 5

vì (2x + 3)² ≥ 0 ∀ x ⇒ 2(2x +3)² + 5 ≥ 5 

A(min) = 5 ⇒ x = - \(\dfrac{3}{2}\)

\(\dfrac{13}{5}\) > 0

\(-\dfrac{7}{9}\) < 0

Vậy \(\dfrac{13}{5}\) > \(-\dfrac{7}{9}\)

độ dài đoạn thẳng AB là : 

AB = OA + OB = 4cm + 2cm = 6cm

độ dài đoạn thẳng NA là : 

N là trung điểm của AB nên

NA = NB = \(\dfrac{AB}{2}=\dfrac{6cm}{2}=3cm\)

độ dài đoạn thẳng MA là : 

M là trung điểm của AO nên 

\(MA=MO=\dfrac{AO}{2}=\dfrac{4cm}{2}=2cm\)

ta có : NA = 3cm và MA = 2cm

⇒ MN = 3cm - 2cm = 1cm

Vậy MN = 1cm

câu a : 

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\\ \left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=\dfrac{7}{4}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}+\dfrac{1}{2}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\\ x-\dfrac{1}{3}=\sqrt{\dfrac{9}{4}}\\ x-\dfrac{1}{3}=\dfrac{3}{2}\\ x=\dfrac{3}{2}+\dfrac{1}{3}\\ x=\dfrac{11}{6}\)

câu b : 

\(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\\ \Rightarrow\left(x-3\right)\cdot\left(x-3\right)=\left(-2\right)\cdot\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ x-3=\sqrt{16}\\ x-3=4\\ x=4+3\\ x=7\)

câu c : 

\(\dfrac{9}{x}=\dfrac{-35}{105}\\ \Rightarrow\left(-35\right)\cdot x=9\cdot105\\ \left(-35\right)\cdot x=945\\ x=945\div\left(-35\right)\\ x=-27\)