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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) 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!
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(6x-42=7y-42\)
\(6x=7y\Leftrightarrow x=\dfrac{7}{6}y\)
\(x=-4:\left(7-6\right).7=-28\)
\(y=-28-4=-24\)
b tương tự
Giải:b)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\) nên \(6\left(x-7\right)=7\left(y-6\right)\)
Do đó \(6x-42=7y-42\) nên \(6x=7y\)
Suy ra \(6x-6y=y\) hay \(6\left(x-y\right)=y\)
Nên 6.(-4) = y
Vậy y = -24, x = \(\dfrac{7.\left(-24\right)}{6}\)= -28
c)
\(\dfrac{x+3}{y+5}=\dfrac{3}{5}\) nên \(5\left(x+3\right)=3\left(y+5\right)\)
Do đó \(5x+15=3y+15\) nên \(5x=3y\)
Suy ra \(5x+5y=3y+5y\)
\(5\left(x+y\right)=8y\)
\(5.16=8y\)
Nên \(y=\dfrac{5.16}{8}=\dfrac{80}{8}=10\)
Vậy y = 10, x = 16 - 10 =6
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\left(\dfrac{1}{2}+\dfrac{1}{3}\right):2\)
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\dfrac{5}{6}:2\)
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\dfrac{5}{12}\)
\(\dfrac{1,3-2,6}{0,363}-x=\dfrac{5}{12}\)
\(\dfrac{-1,3}{0,363}-x=\dfrac{5}{12}\)
\(\dfrac{-1300}{363}-x=\dfrac{5}{12}\)
\(x=\dfrac{-1300}{363}-\dfrac{5}{12}\)
\(x=\dfrac{-1935}{484}\)
tính
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
MONG CÁC BẠN GIÚP ĐỠ XIN CẢM ƠN
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\)
\(=\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{13}{56}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\\ =\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\left(\dfrac{5}{8}\cdot\dfrac{1}{5}\right)-\left(\dfrac{3}{16}\cdot4\right)\\ =\dfrac{6}{7}+\left(\dfrac{1\cdot1}{8\cdot1}\right)-\left(\dfrac{3}{4}\cdot1\right)\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{42}{56}\\ =\dfrac{48+7-42}{56}\\ =\dfrac{13}{56}\)
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)
S=\(\dfrac{1}{1}-\dfrac{1}{4} +...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\)
S=\(\dfrac{1}{1}-\dfrac{1}{100}\)
S=1-\(\dfrac{1}{100}\)
S=\(\dfrac{99}{100}\)
\(M=\dfrac{5^3}{1\cdot6}+\dfrac{5^3}{6\cdot11}+...+\dfrac{5^3}{26\cdot31}\)
\(=5^2\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5^2\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5^2\left(1-\dfrac{1}{31}\right)\)\(=25\cdot\dfrac{30}{31}=\dfrac{750}{31}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)
\(=3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}\right)\)
\(=3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)
\(=1-\dfrac{1}{43}\)
\(=\dfrac{42}{43}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\)
\(=1-\dfrac{1}{43}=\dfrac{42}{43}\)