Tìm số thích hợp điền vào chỗ chấm (...) trong mỗi hình sau, biết:
a) Hình vuông có diện tích: \(\dfrac{81}{16}\)cm vuông . .... cm
b) Hình thoi có diện tích: \(\dfrac{81}{10}\)m2
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a; \(\dfrac{-1}{3}\) \(\times\) \(\dfrac{5}{9}\) = \(\dfrac{-1\times5}{3\times9}\) = \(\dfrac{-5}{27}\)
b; \(\dfrac{-7}{3}\) \(\times\) (-6) = \(\dfrac{-7\times\left(-2\right)}{3}\) \(\times\) 3 = 14
c; \(\dfrac{22}{7}\) : \(\dfrac{11}{14}\) = \(\dfrac{22}{7}\) \(\times\) \(\dfrac{14}{11}\) = \(\dfrac{2\times11}{7}\) \(\times\) \(\dfrac{7\times2}{11}\) = 4
d; \(\dfrac{3}{5}\) + \(\dfrac{4}{7}\) \(\times\) \(\dfrac{14}{6}\) = \(\dfrac{3}{5}\) + \(\dfrac{4}{7}\) \(\times\) \(\dfrac{7\times2}{2\times3}\) = \(\dfrac{3}{5}\) + \(\dfrac{4}{3}\) = \(\dfrac{9}{15}\) + \(\dfrac{20}{15}\) = \(\dfrac{29}{15}\)
a) \(\dfrac{-1}{3}.\dfrac{5}{9}=\dfrac{-1.5}{3.9}=\dfrac{-5}{27}\)
b)\(\dfrac{-7}{3}.\left(-6\right)=\dfrac{\left(-7\right).\left(-6\right)}{3}=\dfrac{42}{3}=14\)
c)\(\dfrac{22}{7}:\dfrac{11}{14}=\dfrac{22}{7}.\dfrac{14}{11}=2.2=4\)
d)\(\dfrac{3}{5}+\dfrac{4}{7}.\dfrac{14}{6}=\dfrac{3}{5}+4.\dfrac{2}{6}=\dfrac{3}{5}+\dfrac{8}{6}=\dfrac{18}{30}+\dfrac{40}{30}=\dfrac{58}{30}=\dfrac{29}{15}\)
Chúc bạn học tốt!
\(C=\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}\)
Ta có: \(4C=2+\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{97}}\)
\(\Rightarrow4C-C=\left(2+\dfrac{1}{2}+\dfrac{2}{2^3}+...+\dfrac{1}{2^{97}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow3C=2-\dfrac{1}{2^{99}}\)
\(\Rightarrow C=\dfrac{2}{3}-\dfrac{1}{3\cdot2^{99}}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{100}}\)
3.D = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) +...+ \(\dfrac{1}{3^{99}}\)
3D - D = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{3^{99}}\) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{100}}\))
2D = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\) - \(\dfrac{1}{3^3}\)-...-\(\dfrac{1}{3^{100}}\)
2D = (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^2}\) - \(\dfrac{1}{3^2}\)) + (\(\dfrac{1}{3^3}\)-\(\dfrac{1}{3^3}\))+...+(\(\dfrac{1}{3^{99}}\)-\(\dfrac{1}{3^{99}}\)) + (1 - \(\dfrac{1}{3^{100}}\))
2D = 0 + 0 + 0 +...+0 + 1 - \(\dfrac{1}{3^{100}}\)
2D = 1 - \(\dfrac{1}{3^{100}}\)
2D = \(\dfrac{3^{100}-1}{3^{100}.2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow B=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
E= 6+8+10+12+14+16+18
Sau khi thay dấu''+'' bằng dấu ''-''
Ta có: E= 6-8-10-12-14-16-18
=> E=-72
Vậy kết quả của bạn Vũ là sai
\(3x+4⋮x+1\\ =>3\left(x+1\right)+1⋮x+1\\ =>1⋮x+1\\ =>x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ x\in\left\{-2;0\right\}\)
Vậy x thuộc {-2;0} thì 3x+4⋮x+1
Gọi độ dài cạnh hình vuông là \(x\) (cm) (đk \(x\) > 0)
Theo bài ta khó: \(x^2\) = \(\dfrac{81}{16}\)
\(\left[{}\begin{matrix}x=\dfrac{9}{4}cm\\x=\dfrac{-9}{4}cm\end{matrix}\right.\)
Vì \(x\) > 0 nên \(x\) = \(\dfrac{9}{4}\) cm
Vậy số cần điền vào chỗ .. là \(\dfrac{9}{4}\)
b; Độ dài đường chéo thứ hai của hình thoi là bao nhiêu em? Thiếu dữ liệu em nhé?
Hình thoi có diện tích 81/10 m vuông nhé