Tính A= \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...........+\frac{1}{512}+\frac{1}{1024}\)
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a) Ta có: \(\widehat{B}=120^o,\widehat{A}=90^o\Rightarrow\widehat{C}+\widehat{D}=360^o-\widehat{A}-\widehat{B}=150^o\)
CO, DO là hai tia phân giác góc C và góc D
=> \(\widehat{C_1}+\widehat{D_1}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)=\frac{1}{2}.150^o=75^o\)
=> \(\widehat{COD}=180^o-\left(\widehat{C_1}+\widehat{D_1}\right)=180^o-75^o=105^o\)
b)
Xét tam giác COD
Ta có: \(\widehat{COD}=180^o-\left(\widehat{C_1}+\widehat{D_1}\right)=180^o-\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)\)
Vì: \(\widehat{C_1}+\widehat{D_1}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)\)
Mặt khác: Xét tứ giác ABCD ta có: \(\widehat{C}+\widehat{D}=360^o-\widehat{A}-\widehat{B}\)
=> \(\widehat{COD}=180^o-\frac{1}{2}\left(360^o-\widehat{A}-\widehat{B}\right)=\frac{1}{2}\widehat{A}+\frac{1}{2}\widehat{B}\)
c) Tương tự ta cũng chứng minh dc:
\(\widehat{BIA}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}\)
=> \(\widehat{COD}+\widehat{BIA}=\frac{1}{2}\widehat{A}+\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}\right)=\frac{1}{2}.360^o=180^o\)
=>\(\widehat{FOE}+\widehat{EIF}=180^o\)
=> \(\widehat{OEI}+\widehat{IFO}=180^o\)
Vậy tứ giác EIF có các góc đối bù nhau!
Ta có BAD + ABC + BCD + CDA = 360 độ
ADC + BCD = 360 - 120 - 90 = 150 độ
=> BCO = OCD = 1/2 BCD
=> ADO = ODC = 1/2 ADC
=> ODC + OCD = 1/2 ODC + 1/2 OCD = ODC+OCD/2
=> ODC + OCD = 150 /2 =75 độ
Mà ODC + OCD +DOC = 180 độ
=> DOC = 180 - 75 = 105 độ
B) COD = 180 - (ODC + OCD)
=> COD = 180 - 1/2ADC + 1/2 BCD
Mà ADC + BCD = 360 - ( BAD + ABC)
COD = 180 - [ 360 - 1/2(BAD + ABC )]
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\(\frac{2}{3\cdot7}+\frac{2}{7\cdot11}+...+\frac{2}{71\cdot75}+\frac{2}{75\cdot79}\)
\(=\frac{2}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{71\cdot75}+\frac{4}{75\cdot79}\right]\)
\(=\frac{2}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{75}-\frac{1}{79}\right]\)
\(=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{79}\right]=\frac{38}{237}\)
\(\frac{2}{3\cdot7}+\frac{2}{7\cdot11}+...+\frac{2}{71\cdot75}+\frac{2}{75\cdot79}\)
\(=\frac{1}{2}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{71\cdot75}+\frac{4}{75\cdot79}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{71}-\frac{1}{75}+\frac{1}{75}-\frac{1}{79}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{79}\right)\)
\(=\frac{1}{2}\cdot\frac{76}{237}\)
\(=\frac{38}{237}\)
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VP=\(A^2X^2+B^2Y^2+C^2Z^2+A^2Y^2+B^2X^2+A^2Z^2+C^2X^2+B^2Z^2+C^2Y^2\)
=\(A^2\left(X^2+Y^2+Z^2\right)+B^2\left(X^2+Y^2+Z^2\right)+C^2\left(X^2+Y^2+Z^2\right)\)
=\(\left(X^2+Y^2+Z^2\right)\left(A^2+B^2+C^2\right)\)
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\(\frac{1-1+2+9+8\cdot8}{x-36.12+5}=3615\)
\(\Rightarrow\frac{1-1+2+9+64}{x-432+5}=3615\)
\(\Rightarrow\frac{75}{x-427}=3615\)
\(\Rightarrow\left(x-427\right)3615=75\)
\(\Rightarrow3615x-1543605=75\)
\(\Rightarrow3615x=1543680\)
\(\Rightarrow x=\frac{1543680}{3615}=\frac{102912}{241}\)
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1. a) M nằm giữa A và B nên AM + MB = AB
=> AM = AB - MB = 10 - 4 = 6 (cm)
Ta có: AN = NM = AM/2 = 6/2 = 3 (cm)
M nằm giữa B và N nên BM + MN = BN
=> BN = 4 + 3 = 7 (cm)
b) Ta có: BI = IM = BM/2 = 4/2 = 2 (cm)
M nằm giữa N và I nên IM + MN = NI
=> NI = 2 + 3 = 5 (cm)
2.
A nằm giữa O và B (OA < OB) nên OA + AB = OB
=> AB = OB - OA = 10 - 5 = 5 (cm)
=> OA = AB = 5 (cm)
=> A là trung điểm của OB
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\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)