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

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

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

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 = - \(\dfrac{3}{2}\)

Tìm GTNN của biểu thức (2x+5)⁴+3

\(\dfrac{13}{5}\) > 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 : 

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

câu a : 

\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)

câu b : 

\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)

câu c : 

\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)

câu d : 

\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)

câu e : 

\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)

câu g : 

\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)

câu h : 

\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)

\(6xy-3x+2y=13\)

\(\Leftrightarrow6xy-3x+2y-1=12\)

\(\Leftrightarrow3x\left(2y-1\right)+2y-1=12\)

\(\Leftrightarrow\left(3x+1\right)\left(2y-1\right)=12\)

Mặt khác \(2y-1\) luôn lẻ nên ta chỉ cần xét các cặp ước \(\left(12;1\right);\left(4;3\right);\left(-12;-1\right);\left(-4;-3\right)\)

Vậy có đúng 1 cặp số nguyên thỏa mãn đề bài là \(\left(x;y\right)=\left(1;2\right)\)

đổi : \(1\dfrac{1}{5}\) giờ = \(\dfrac{6}{5}\) giờ; 70 phút = \(\dfrac{7}{6}\)  giờ

vận tốc xe taxi là : 

\(100\div\dfrac{6}{5}=100\times\dfrac{5}{6}\approx83\left(km\text{/}h\right)\)

vận tốc xe tải là : 

\(100\div\dfrac{7}{6}=100\times\dfrac{6}{7}\approx86\left(km\text{/}h\right)\)

vì 86km/h > 83km/h nên => xe tải nhanh hơn xe taxi

nhầm