\(\dfrac{10^2+11^2+12^2}{13^2+14^2}\)
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\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
Ta có: \(\dfrac{3}{10}>\dfrac{3}{15}\)
\(\dfrac{3}{11}>\dfrac{3}{15}\)
\(\dfrac{3}{12}>\dfrac{3}{15}\)
\(\dfrac{3}{13}>\dfrac{3}{15}\)
\(\dfrac{3}{14}>\dfrac{3}{15}\)
Do đó: \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}=1\)
hay 1<S(1)
Ta có: \(\dfrac{3}{11}< \dfrac{3}{10}\)
\(\dfrac{3}{12}< \dfrac{3}{10}\)
\(\dfrac{3}{13}< \dfrac{3}{10}\)
\(\dfrac{3}{14}< \dfrac{3}{10}\)
Do đó: \(\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{12}{10}\)
\(\Leftrightarrow S< \dfrac{15}{10}=\dfrac{3}{2}< 2\)(2)
Từ (1) và (2) suy ra 1<S<2(đpcm)
a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Vậy \(A=\dfrac{3}{11}\)
b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)
Vậy \(B=6\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)
Lời giải:
\(\frac{10^2+11^2+12^2}{13^2+14^2}=\frac{365}{365}=1\)