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NN
1
TN
12 tháng 6 2016
A=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...\)
A-\(\frac{1}{5}=\frac{1}{5^2}+\frac{1}{5^3}+...\)
5 -\(\left(A-\frac{1}{5}\right)=5.\left(\frac{1}{5^2}+\frac{1}{5^3}+...1\right)\)
10.\(\left(A-\frac{1}{5}\right)=A\)
\(10A-5=A\)
\(10.A-A=5\)
\(9A=5\)
\(A=\frac{5}{9}\)
T hk pk dung hay sai nha, ma m cu lam ik, co j len t sua
20 tháng 8 2023
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
20 tháng 8 2023
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
S
31 tháng 8 2017
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
28 tháng 7 2019
a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha
.
HP
16 tháng 4 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\Rightarrow x+1=2011\Rightarrow x=2010\)
Vậy x=2010
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-99}{100}\)(9 thừa số)
\(=-\frac{3.8.15...99}{4.9.16...100}=\frac{1.3.2.4.3.5...9.11}{2.2.3.3.4.4...10.10}=\frac{\left(1.2.3....9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right).\left(2.3.4...10\right)}=\frac{11}{10.2}=\frac{11}{20}\)
Đặt \(A=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....................\left(\frac{1}{100}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{4}-\frac{4}{4}\right).\left(\frac{1}{9}-\frac{9}{9}\right)..................\left(\frac{1}{100}-\frac{100}{100}\right)\)
\(\Rightarrow A=\frac{-3}{4}.\frac{-8}{9}................\frac{-99}{100}\)
\(\Rightarrow A=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}..............\frac{-9.11}{10.10}\)
\(\Rightarrow A=-\frac{\left(1.2..................9\right).\left(3.4.............11\right)}{\left(2.3...........10\right).\left(2.3...............10\right)}=\frac{1.11}{10.2}=\frac{11}{20}\)
Vậy \(A=\frac{11}{20}\)
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