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Có: \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{31-26}{26.31}\right)\)
\(=5.\left(\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+...+\frac{31}{26.31}-\frac{26}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}=\frac{150}{31}>\frac{31}{31}=1\)
\(\Rightarrow A>1\)
Ta có: A=\(\frac{5^2}{1.6}\)+\(\frac{5^2}{6.11}\)+...+\(\frac{5^2}{26.31}\)
=5.(\(\frac{5}{1.6}\)+\(\frac{5}{6.11}\)+...+\(\frac{5}{26.31}\))
=5.(1-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)+...+\(\frac{1}{26}\)-\(\frac{1}{30}\))
=5.(1-\(\frac{1}{30}\))
=5.\(\frac{29}{30}\)
=\(\frac{29}{6}\)>1
Hay A>1
=> đpcm
Bài 2:
Ta thấy: 52 > 4.5
62 > 5.6
72 > 6.7
....
20172 > 2016.2017
\(\Rightarrow\)\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
....
\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)
Cộng vế với nhau, ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}\)( vì \(\frac{1}{2017}>0\))
k giúp mik ✅
\(a,3^6:3^5=3^{6-5}=3\\ b,5^7:5^5=5^{7-5}=5^2=25\\ c,14^5:2^3:7^4=\left(2^5:2^3\right)\cdot\left(7^5:7^4\right)=2^2\cdot7=28\\ d,5^4-2\cdot5^3=5^3\left(5-2\right)=3\cdot5^3=375\)
a) 3^6 : 3^5 = 729 : 243 = 3
b) 5^7 : 5^5 = 78125 : 3125 = 25
c) 14^5 : 2^3 : 7^4 = 537824 : 8 : 2401 = 89
d) 5^4 - 2 * 5^3 = 625 - 2 * 125 = 625 - 250 = 375
\(a,3^6:3^5=3^{6-5}=3^1=3\\ b,5^7:5^5=5^{7-5}=5^2=25\\ c,14^5:2^3:7^4=\left(2^5:2^3\right).\left(7^5:7^4\right)=2^{5-3}.7^{5-4}=2^2.7^1=4.7=28\\ d,5^4-2.5^3=5.5^3-2.5^3=\left(5-2\right).5^3=3.5^3=3.125=375\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5^2\cdot2\)
\(=5\cdot\left[\left(85-5\right):8+90\right]-25\cdot2\)
\(=5\cdot\left(10+90\right)-50\)
=450
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
Sửa đề \(\dfrac{5^3}{1.6}\rightarrow\dfrac{5^2}{1.6}\)
Giải:
\(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(\dfrac{1}{1}-\dfrac{1}{31}\right)\)
\(=5.\dfrac{30}{31}\)
\(=\dfrac{150}{31}\)