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\(B=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\frac{2+8+18+32+50}{12+48+108+192+300}\\ =\frac{110}{660}=\frac{1}{6}\)
B=\(\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\)
B=\(\dfrac{1.2\left(4+3^2+2.8+5^2\right)}{3.4\left(2^2+3^2+2.8+5^2\right)}\)
B=\(\dfrac{1.2}{3.4}=\dfrac{1}{6}\)
\(=\frac{1.2\left(1+2.2+3.3+4.4+5.5\right)}{3.4\left(1+2.2+3.3+4.4+5.5\right)}\)
\(=\frac{1.2}{3.4}=\frac{1}{6}\)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.2+2.3.4.2+3.3.6.2+4.3.8.2+5.3.10.2}\)=\(\frac{1}{3.2+3.2+3.2+3.2+3.2}\)=\(\frac{1}{30}\)
vậy bt trên =\(\frac{1}{30}\)
tk nhé
Ta có:
\(D=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\\ =\dfrac{1.2\left(4+3^2+2.8+5^2\right)}{3.4\left(2^2+3^2+2.8+5^2\right)}\\ =\dfrac{1.2}{3.4}=\dfrac{1}{6}\)
Vậy \(D=\dfrac{1}{6}\)
Theo đề bài ta có :
\(A=\frac{n+1}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow2\left(n+1\right)=n-1\)
\(\Leftrightarrow2n+2=n-1\)
\(\Leftrightarrow2n-n=-1-2\)
\(\Rightarrow n=-3\)
Vậy với n = - 3 thì A = \(\frac{1}{2}\)
\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\dfrac{1.2\left(1^2+2^2+3^2+2^4+2^5\right)}{3.4\left(1^2+2^2+2^3+2^4+2^5\right)}=\dfrac{2}{12}=\dfrac{1}{6}\)