Cho: \(\left(x+\frac{1}{1\cdot2}\right)\)+\(\left(x+\frac{1}{2\cdot3}\right)\)+...+\(\left(x+\frac{1}{2018\cdot2019}\right)\)=2019x
Tìm x
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giải pt
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.......+\frac{1}{2005.2006.2007}\right)x=\left(1\cdot2+2\cdot3+.....+2006.2007\right)\)
(\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{9\cdot10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{9.10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(1-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}x-\frac{9}{10}+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(\frac{9}{10}x+\frac{1}{10}x\right)-\frac{9}{10}=x-\frac{9}{10}\)
\(\Rightarrow x-\frac{9}{10}=x-\frac{9}{10}\)
\(\Rightarrow x\inℝ\)
Vậy \(x\inℝ\)
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Tính giá trị biểu thức
\(A=\left|x-\frac{1}{1\cdot2}\right|+\left|x-\frac{1}{2\cdot3}\right|+\left|x-\frac{1}{3\cdot4}\right|+...+\left|x-\frac{1}{100\cdot101}\right|+100x\)
Với \(x< 0\)
\(A=\frac{1}{1.2}-x+\frac{1}{2.3}-x+...+\frac{1}{100.101}-x+100x\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}-100x+100x\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Tìm x, biết :a, left(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{98cdot99cdot100}right)x-3;b, left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}}{frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right)xfrac{-1}{5}.c,left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}+2000}{1+frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right):xfrac{-2001}{2002}.
Đọc tiếp
Tìm x, biết :
a, \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\right)x=-3\);
b, \(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right)x=\frac{-1}{5}\).
c,\(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right):x=\frac{-2001}{2002}\).
Bài 1: Tínha. left(1+frac{1}{1cdot3}right)cdotleft(1+frac{1}{2cdot4}right)cdotleft(1+frac{1}{3cdot5}right)+left(1+frac{1}{4cdot6}right).....left(1+frac{1}{99cdot101}right)b. left[sqrt{0,64}+sqrt{0,0001}-sqrt{left(-0,5right)^2}right]divleft[3cdotsqrt{left(0,04right)^2}-sqrt{left(-2right)^4}right]c. frac{5.4^{15}cdot9^9-4.3^{20}cdot8^9}{5cdot2^9cdot6^{19}-7cdot2^{29}cdot27^6}-frac{2^{19}cdot6^{15}-7cdot6^{10}cdot2^{20}cdot3^6}{9cdot6^{19}cdot2^9-4cdot3^{17}cdot2^{26}}+0,left(6right)Bài 2: Tìm x, y...
Đọc tiếp
Bài 1: Tính
a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)
b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)
c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)
Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)
b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\)
c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)
d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)
e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)
Bài 3: Chứng minh rằng
a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)
b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)
Bài 4:
a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)
b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)
c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
5 tháng 3 2020 lúc 15:24
Tính giá trị của biểu thức:
\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2017\cdot2019}\right)\)
A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)
Vậy \(A=\frac{2018}{2019}\)
Chúc bn học tốt
\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)
\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)
\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)
\(=2018.\frac{2}{2019}\)
\(=\frac{4036}{2019}\)
\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)
\(A=\frac{2018}{2019}\)
Tìm số tự nhiên x biết:
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
Tính tổng :a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính tổng :
a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}
b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}
c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}
d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}
e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
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