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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) Để phân số \(\dfrac{12}{n}\) có giá trị nguyên thì :
\(12⋮n\)
\(\Leftrightarrow n\inƯ\left(12\right)\)
\(\Leftrightarrow n\in\left\{-1;1;-12;12;-2;2;-6;6;-3;3;-4;4\right\}\)
Vậy \(n\in\left\{-1;1;-12;12;-2;2-6;6;-3;3;-4;4\right\}\) là giá trị cần tìm
b) Để phân số \(\dfrac{15}{n-2}\) có giá trị nguyên thì :
\(15⋮n-2\)
\(\Leftrightarrow x-2\inƯ\left(15\right)\)
Tới đây tự lập bảng zồi làm típ!
c) Để phân số \(\dfrac{8}{n+1}\) có giá trị nguyên thì :
\(8⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(8\right)\)
Lập bảng rồi làm nhs!
a) 4.(-5)2+(-2)3.25
= 4.25+(-8).25
=25.[4+(-8)]
=25.(-4)
=-100
b)\(15\dfrac{3}{7}-\left(\dfrac{7}{15}+9\dfrac{4}{7}\right)\)
= \(15\dfrac{3}{7}-\dfrac{7}{15}-9\dfrac{4}{7}\)
= \(\left(15\dfrac{3}{7}-9\dfrac{4}{7}\right)-\dfrac{7}{15}\)
=\(\left(14\dfrac{10}{7}-9\dfrac{4}{7}\right)-\dfrac{7}{15}\)
=\(5\dfrac{1}{7}-\dfrac{7}{15}\)
=\(\dfrac{36}{7}-\dfrac{7}{15}\)
=\(\dfrac{540}{105}-\dfrac{49}{105}\)
=\(\dfrac{491}{105}\)
\(\)a) 4.(-5)2+(-2)3.25
\(=4.5^2+\left(-2\right)^3.25\)
\(=4.25+\left(-8\right).25\)
\(=100+\left(-200\right)\)
\(=-100\)
b) \(15\dfrac{3}{7}-\left(\dfrac{7}{15}+9\dfrac{4}{7}\right)\)
\(=\dfrac{108}{7}-\left(\dfrac{7}{15}+\dfrac{67}{7}\right)\)
\(=\dfrac{108}{7}-\dfrac{1054}{105}\)
\(=\dfrac{566}{105}\)
Đặt A = \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
\(\Rightarrow2A=\)\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{15}=\dfrac{14}{15}\)
\(\Rightarrow A=\dfrac{14}{15}:2=\dfrac{7}{15}\)
A =\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
A = \(\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-...-\left(\dfrac{1}{65}-\dfrac{1}{65}\right)-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)