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a) \(\dfrac{x-1}{x^2-4}=\dfrac{3}{2-x}\)
\(\Leftrightarrow\dfrac{x-1}{\left(x-2\right)\left(x+2\right)}=-\dfrac{3}{\left(x-2\right)\left(x+2\right)}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow x-1=-3\)
\(\Leftrightarrow x=1-3=-2\)
Vậy: \(x=-2\)
b) \(\dfrac{1}{x-1}-\dfrac{7}{x-2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\left(-\dfrac{7}{2-x}\right)=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2-x}{\left(x-1\right)\left(2-x\right)}+\dfrac{7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Rightarrow2-x+7x-7=1\)
\(\Leftrightarrow-x+7x=1-2+7=6\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
c) \(\dfrac{2x+3}{2x-3}-\dfrac{3}{4x-6}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{2x+3}{2x-3}-\dfrac{3}{2\left(2x-3\right)}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{10\left(2x+3\right)}{10\left(2x-3\right)}-\dfrac{3.5}{10\left(2x-3\right)}=\dfrac{4\left(2x-3\right)}{10\left(2x-3\right)}\)
\(ĐKXĐ:x\ne\dfrac{3}{2}\)
\(\Leftrightarrow10\left(2x+3\right)-15=4\left(2x-3\right)\)
\(\Leftrightarrow20x+30-15=8x-12\)
\(\Leftrightarrow20x-8x=15-12-30\)
\(\Leftrightarrow12x=-27\)
\(\Leftrightarrow x=-\dfrac{27}{12}=-\dfrac{9}{4}\)
Vậy: \(x=-\dfrac{9}{4}\)
d) \(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)
\(\Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}=\dfrac{x+60}{43}-\dfrac{x+60}{45}\)
\(\Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\)
\(\Leftrightarrow x+60=0\) vì \(\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\)
\(\Leftrightarrow x=-60\)
Vậy: \(x=-60\)
_Good luck to you_
\(a.C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}=\dfrac{x^{28}+x^{24}+...+x^8+x^4+1}{x^3\left(x^{28}+x^{24}+...+x^8+x^4+1\right)}=\dfrac{1}{x^3}\) Tại x = 2015 thì : \(C=\dfrac{1}{x^3}=\dfrac{1}{2015^3}\)
\(b.F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
\(3F=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{2011.2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{2011.2012.2013}-\dfrac{1}{2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2012.2013.2014}\)
Tới đây dễ rồi , bạn tự tính nốt .
Vì làm vậy để triệt tiêu dần mà ( dang bài kiểu ... này thường là phải triệt tiêu ) Triệu Tử Dương
Hình như thiếu mũ 2007 -.- Sửa luôn nhóe :)
Trước hết ta tính tổng sau, với các số tự nhiên a, n đều lớn hơn 1.
\(S_n=\dfrac{1}{a}+\dfrac{1}{a^2}+...+\dfrac{1}{a^n}\)
Ta có: \(\left(a-1\right)S_n=aS_n-S_n\)
\(=\left(1+\dfrac{1}{a}+\dfrac{1}{a^2}+...+\dfrac{1}{a^{n-1}}\right)-\left(\dfrac{1}{a}+\dfrac{1}{a^2}+...+\dfrac{1}{a^{n-1}}+\dfrac{1}{a^n}\right)\)\(=1-\dfrac{1}{a^n}< 1\Rightarrow S_n< \dfrac{1}{a-1}\left(1\right)\)
Áp dụng BĐT ( 1 ) cho a = 2008 và mọi n = 2,3, ..., 2004 ta được:
\(B=\dfrac{1}{2008}+\left(\dfrac{1}{2008}+\dfrac{1}{2008^2}\right)^2+...+\left(\dfrac{1}{2008}+\dfrac{1}{2008^2}+...+\dfrac{1}{2008^{2007}}\right)^{2007}< \dfrac{1}{2007}+\left(\dfrac{1}{2007}\right)^2+...+\left(\dfrac{1}{2007}\right)^{2007}\left(2\right)\)
Lại áp dụng BĐT ( 1 ) cho a = 2007 và n = 2007, ta được:
\(\dfrac{1}{2007}+\dfrac{1}{2007^2}+...+\dfrac{1}{2007^{2007}}< \dfrac{1}{2006}=A\left(3\right)\)
Từ ( 2 ) và ( 3 ) => B < A.
Bài 1 :
a) +) \(\dfrac{1}{8}\cdot16^n=2^n\)
\(\Leftrightarrow\dfrac{1}{8}=\dfrac{2^n}{16^n}\)
\(\Rightarrow\dfrac{1}{8}=\dfrac{1}{8}^n\)
Vậy n = 1.
+) \(27< 3^n< 243\)
\(\Leftrightarrow3^3< 3^n< 3^5\)
Vậy n = 4.
Bài 2 : \(\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}=-\dfrac{45}{28}\)
Bài 2 :
chưa hiểu: @Duc Minh
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)=\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}+...-\dfrac{1}{49}\right)\)
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
b,\(\dfrac{4}{9}x^2+4x+9=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.3+3^2=\left(\dfrac{2}{3}x+3\right)^2\)
c, \(x^3+9x^2+27x+27=x^3+3.x^2.3+3.x.3^2+3^3=\left(x+3\right)^3\)
d, \(\dfrac{1}{8}-\dfrac{3}{4}x+\dfrac{3}{2}x^2-x^3=\left(\dfrac{1}{2}\right)^3-3.\left(\dfrac{1}{2}\right)^2.x+3.\dfrac{1}{2}.x^2-x^3=\left(\dfrac{1}{2}-x\right)^3\)
TK MIK 
$A=\dfrac{2018.2017-1}{2016.2018+2017}$
$=>A={2018.2016+2018-1}{2016.2018+2017}$
$=>A={2018.2016+2017}{2016.2018+2017}$
$=>A=1$
\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)
\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)
Chúc bạn học tốt!!!