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\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
a, (x2 - 5)(x2 - 24) < 0
=> x2 - 5 và x2 - 24 trái dấu
Mà x2 - 5 > x2 - 24 => \(\hept{\begin{cases}x^2-5>0\\x^2-24>0\end{cases}\Rightarrow5< x^2< 24}\)
Vì x \(\in\)Z nên x2 = 9;16
+) x2 = 9 => x = 3 hoặc x = -3
+) x2 = 16 => x = 4 hoặc x = -4
Vậy...
b,
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x + 1 = 0 => x = 0 - 1 => x = -1
\(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Rightarrow\left(\frac{x+1}{14}+1\right)+\left(\frac{x+2}{13}+1\right)=\left(\frac{x+3}{12}+1\right)+\left(\frac{x+4}{11}+1\right)\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Rightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Mà \(\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)\ne0\)
=> x + 15 = 0 => x = 0 - 15 => x = -15
Xét Sn = 1+2+3+4+...+n (1)
=> Sn= n+(n-1)+...+2+1 (2)
Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1
Lấy (1);(2) và chú ý trên ta có:
2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1) (vì n số hạng giống nhau)
=> Sn= n(n+1)/2 => Sn/n = (n+1)/2
=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n
P= 1+3/2+4/2+5/2+...+(n+1)/2
P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2
P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n2+3n
Bài toán chỉ đến S2016/2016 (tức n=2016)
Vậy S= 20162+3.2016=2016.(2016+3)=2016.2019=4070304
E = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 + 3) + 1/4.(1 + 2 + 3 + 4) + ... + 2016.(1 + 2 + 3 + ... + 2016)
E = 1 + 1/2.(1 + 2).2:2 + 1/3.(1 + 3).3:2 + 1/4.(1 + 4).4:2 + ... + 2016.(1 + 2016).2016:2
E = 2/2 + 3/2 + 4/2 + 5/2 + ... + 2017/2
E = 2+3+4+5+...+2017/2
E = (2 + 2017).2016/2
E = 2019.1008
E = 2 035 152
Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow B< 1-\frac{1}{100}< 1\)
\(\Rightarrow B< 1\Rightarrow B< A\)
Vậy B<A