Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
A, \(\dfrac{36}{288}+\dfrac{9}{288}+\dfrac{4}{288}=\dfrac{49}{288}\)
b,\(\dfrac{3}{5}\)
c,\(\dfrac{34}{35}-\dfrac{15}{35}-\dfrac{14}{35}=\dfrac{1}{7}\)
d,\(\dfrac{1}{2}-\dfrac{6}{7}=\dfrac{7}{14}-\dfrac{12}{14}=\dfrac{-5}{14}\)
a.\(\dfrac{15}{17}.\left(\dfrac{4}{5}+\dfrac{1}{5}\right)=\dfrac{15}{17}.1=\dfrac{15}{17}\)
b,\(\dfrac{1}{2}:\left(\dfrac{29}{12}-\dfrac{5}{12}\right)=\dfrac{1}{2}:\dfrac{24}{12}=\dfrac{1}{4}\)
a) Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in Z\right)\)
\(\Rightarrow B=\dfrac{5^{12}+2}{5^{13}+2}< 1\)
\(B< \dfrac{5^{12}+2+48}{5^{13}+2+48}\Rightarrow B< \dfrac{5^{12}+50}{5^{13}+50}\Rightarrow B< \dfrac{5^2\left(5^{10}+2\right)}{5^2\left(5^{11}+2\right)}\Rightarrow B< \dfrac{5^{10}+2}{5^{11}+2}=A\)\(B< A\)
Ta có:
\(\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d};\dfrac{b}{a+b+d}< \dfrac{b+c}{a+b+c+d}\)
\(\dfrac{c}{b+c+d}< \dfrac{c+a}{a+b+c+d};\dfrac{d}{a+c+d}< \dfrac{b+d}{a+b+c+d}\)
Cộng theo vế các BĐT trên ta có:
\(P< \dfrac{a+d}{a+b+c+d}+\dfrac{b+c}{a+b+c+d}+\dfrac{c+a}{a+b+c+d}+\dfrac{b+d}{a+b+c+d}=\dfrac{2\left(a+b+c+d\right)}{a+b+c+d}=2\left(1\right)\)
Lại có:
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d};\dfrac{b}{a+b+d}>\dfrac{b}{a+b+c+d}\)
\(\dfrac{c}{b+c+d}>\dfrac{c}{a+b+c+d};\dfrac{d}{a+c+d}>\dfrac{d}{a+b+c+d}\)
Cộng theo vế các BĐT trên có:
\(P>\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}=\dfrac{a+b+c+d}{a+b+c+d}=1\left(2\right)\)
Từ \((1);(2)\) ta thu được ĐPCM
Câu 1:
\(A\in Z\Rightarrow6n-1⋮3n+2\)
\(\Rightarrow6n+4-5⋮3n+2\)
\(\Rightarrow2\left(3n+2\right)-5⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
đến đây tự lm nốt nhé
1. Để A có giá trị nguyên thì \(6n-1⋮3n+2\)
Ta có: \(\left\{{}\begin{matrix}6n-1⋮3n+2\\3n+2⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\2\left(3n+2\right)⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n-1+5⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left(6n-1+5\right)-\left(6n-1\right)⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
\(\Rightarrow3n+2\inƯ\left(5\right)\)
\(\Rightarrow3n+2\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow3n\in\left\{-7;\pm3;-1;\right\}\)
\(\Rightarrow n\in\left\{\pm1\right\}\)
Vậy để \(A\in Z\) thì n nhận các giá trị là: \(\pm1\)