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\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)
\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)
\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
...
\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)
Vậy ...
b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
...
\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)
Vậy ...
áp dụng hằng đẳng thức (a+b)(a-b)=a^2-b^2 Minh Hoang Hai
a) Ta có: \(P=\left(\dfrac{x^2-1}{x^4-x^2+1}+\dfrac{2}{x^6+1}-\dfrac{1}{x^2+1}\right)\cdot\left(x^2-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)
\(=\left(\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\dfrac{2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}-\dfrac{x^4-x^2+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\cdot\left(\dfrac{x^2\left(x^4+x^2+1\right)}{x^4+x^2+1}-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)
\(=\dfrac{x^4-1+2-x^4+x^2-1}{\left(x^2+1\right)\cdot\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+x^4+x^2-x^4-x^2+1}{x^4+x^2+1}\)
\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+1}{x^4+x^2+1}\)
\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^4+x^2+1}\)
\(=\dfrac{x^2}{x^4+x^2+1}\)
a: |x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2(nhận) hoặc x=4(loại)
Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)
b: ĐKXĐ: x<>4; x<>-4
\(B=\dfrac{-\left(x+4\right)}{x-4}+\dfrac{x-4}{x+4}-\dfrac{4x^2}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{-x^2-8x-16+x^2-8x+16-4x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{-4x^2-16x}{\left(x-4\right)\left(x+4\right)}\)
=-4x/x-4
c: A+B
=-4x/x-4+x^2+4/x-4
=(x-2)^2/(x-4)
A+B>0
=>x-4>0
=>x>4
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
A= \(\dfrac{1^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
= \(\dfrac{1}{1\cdot3}\cdot\dfrac{3^2}{3\cdot5}\cdot\dfrac{5^2}{5\cdot7}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
=\(\dfrac{1}{n+2}\)
B = \(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}=\dfrac{16}{1-x^{16}}\)
a: \(A=\dfrac{x^2+1}{x}+\dfrac{x^3-1}{x^2-x}+\dfrac{x^4-x^3+x-1}{x-x^3}\)
\(=\dfrac{x^2+1}{x}+\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}-\dfrac{x^3\left(x-1\right)+\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1}{x}+\dfrac{x^2+x+1}{x}-\dfrac{\left(x-1\right)\left(x^3+1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1+x^2+x+1}{x}-\dfrac{x^2-x+1}{x}\)
\(=\dfrac{2x^2+x+2-x^2+x-1}{x}=\dfrac{x^2+2x+1}{x}=\dfrac{\left(x+1\right)^2}{x}\)
b: \(x^2+x=12\)
=>\(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)
Thay x=3 vào A, ta được:
\(A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)
Khi x=-4 thì \(A=\dfrac{\left(-4+1\right)^2}{-4}=\dfrac{9}{-4}=-\dfrac{9}{4}\)
c: \(A-4=\dfrac{\left(x+1\right)^2}{x}-4\)
\(=\dfrac{\left(x+1\right)^2-4x}{x}\)
\(=\dfrac{x^2+2x+1-4x}{x}=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}\)>0 với mọi x>0
=>A>4
Lời giải:
Ta có: \(4+(2n-1)^4=[(2n-1)^2+2]^2-[2(2n-1)]^2\)
\(=[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]\)
\(\Rightarrow \frac{2n-1}{4+(2n-1)^4}=\frac{2n-1}{[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]}\)
\(=\frac{1}{4}\left(\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)}\right)\)
Do đó:
\(\frac{1}{4+1^4}=\frac{1}{4}(1-\frac{1}{5})\)
\(\frac{3}{4+3^4}=\frac{1}{4}(\frac{1}{5}-\frac{1}{17})\)
\(\frac{5}{4+5^4}=\frac{1}{4}(\frac{1}{17}-\frac{1}{37})\)
......
Do đó:
\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+(2n-1)^4}=\frac{1}{4}(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{17}+...+\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)})\)
\(=\frac{1}{4}(1-\frac{1}{(2n-1)^2+2+2(2n-1)})=\frac{1}{4}(1-\frac{1}{(2n-1+1)^2+1})\)
\(=\frac{1}{4}(1-\frac{1}{4n^2+1})=\frac{n^2}{4n^2+1}\)
Ta có đpcm.
n=1 ; \(\dfrac{1}{4+1^4}=\dfrac{1}{5}=\dfrac{1^2}{4.^2+1}=\dfrac{1}{5};dung\)
giả sử n =k đúng \(\Leftrightarrow S=\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\) (*)
cần c/m đúng n =k+1 ;
c/m
với n=k+1
\(S=\left(\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}\right)+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
từ (*) =>\(S=\dfrac{k^2}{4k^2+1}+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
\(k+1=t\Leftrightarrow k=t-1\)
\(S=\dfrac{t^2-2t+1}{4\left(t^2-2t+1\right)+1}+\dfrac{2t-1}{4+\left(2t-1\right)^4}\)
\(S=\dfrac{t^2-2t+2}{4t^2-8t+5}+\dfrac{2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{\left(t^2-2t+1\right)\left(4t^2+1\right)+2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}\)\(S=\dfrac{t^2\left(4t^2-8t+5\right)}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{t^2}{\left(4t^2+1\right)}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)
Vậy tổng trên đúng với k +1
theo Quy nạp ta có dpcm
câu a: 14a + 6b = 84 + ab
<=> 14a + 6b - 84 - ab =0
<=> (14a -84) + (6b -ab)=0
<=> 14( a- 6) - b(a-6)=0
<=> (a - 6)(14-b) = 0
Vậy a=6, b=14
Đặt \(A=\dfrac{n}{4+n^4}\)
\(=\dfrac{n}{n^4+4n^2+4-4n^2}\)
\(=\dfrac{n}{\left(n^2+2\right)^2-\left(2n\right)^2}\)
\(=\dfrac{n}{\left(n^2+2-2n\right)\left(n^2+2+2n\right)}\)
\(\Rightarrow4A=\dfrac{4n}{\left(n^2-2n+2\right)\left(n^2+2n+2\right)}\)
\(=\dfrac{1}{n^2-2n+2}-\dfrac{1}{n^2+2n+2}\)
Đặt \(P=\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2n-1}{4+\left(2n-1\right)^4}\)
\(\Rightarrow4P=\dfrac{4}{4+1^4}+\dfrac{12}{4+3^4}+...+\dfrac{4\left(2n-1\right)}{4+\left(2n-1\right)^4}\)
\(=\dfrac{1}{1^2-2\times1+2}-\dfrac{1}{1^2+2\times1+2}\)
\(+\dfrac{1}{3^2-2\times3+2}-\dfrac{1}{3^2+2\times3+2}+...+\)
\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{17}+...+\)
\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{4n^2-4n+1+4n-2+2}\)
\(=1-\dfrac{1}{4n^2+1}\)
\(\Rightarrow P=\dfrac{1}{4}-\dfrac{1}{4\left(4n^2+1\right)}\)