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

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!

Đề bài yêu cầu gì vậy bạn?

Đề bài là gì bạn ?

\(C=\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}\)

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

20,3

20,3

5/6 của 2/5 là:

2/5 x 5/6 = 1/3

1/

\(3x+4⋮x+1\\ =>3\left(x+1\right)+1⋮x+1\\ =>1⋮x+1\\ =>x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ x\in\left\{-2;0\right\}\)

Vậy x thuộc {-2;0} thì 3x+4⋮x+1

Mẫu giáo?

cứ làm giúp tớ vs :'<