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\(C=\dfrac{-1}{5}+\left(\dfrac{1}{-5}\right)^2+\left(-\dfrac{1}{5}\right)^3+...+\left(-\dfrac{1}{5}\right)^{99}\)
=>\(5\cdot C=-1+\left(-\dfrac{1}{5}\right)+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{98}\)
=>\(5\cdot C-C=\left(-1\right)-\left(-\dfrac{1}{5}\right)^{99}\)
=>\(4C=-1+\dfrac{1}{5^{99}}=\dfrac{-5^{99}+1}{5^{99}}\)
=>\(C=\dfrac{-5^{99}+1}{4\cdot5^{99}}\)
(x-3y)^2006+(y+4)^2008=0
=>x-3y=0 và y+4=0
=>x=3y và y=-4
=>x=3*(-4)=-12 và y=-4
Chứng minh rằng:
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
Ta có:
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\\ =\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100}< 2\)
Chứng minh rằng:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
Dễ thấy \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\) do đó:
\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Bài 1 :
\(A=\dfrac{n+1}{n+2}\) có giá trị nguyên âm, dương khi
\(n+1⋮n+2\)
\(\Rightarrow n+1-\left(n+2\right)⋮n+2\)
\(\Rightarrow n+1-n-2⋮n+2\)
\(\Rightarrow-1⋮n+2\)
\(\Rightarrow n+2\in\left\{-1;1\right\}\)
\(\Rightarrow n\in\left\{-3;-1\right\}\left(n\in Z\right)\)
Bài 2 :
\(1+\left(-\dfrac{1}{60}\right)+\dfrac{19}{120}< \dfrac{x}{36}+\left(-\dfrac{1}{60}\right)< \dfrac{58}{90}+\dfrac{59}{72}+\left(-\dfrac{1}{60}\right)\)
\(\Rightarrow1+\dfrac{19}{120}< \dfrac{x}{36}< \dfrac{58}{90}+\dfrac{59}{72}\)
\(\Rightarrow\dfrac{139}{120}< \dfrac{x}{36}< \dfrac{232}{360}+\dfrac{295}{360}\)
\(\Rightarrow\dfrac{417}{360}< \dfrac{10x}{360}< \dfrac{527}{360}\)
\(\Rightarrow417< 10x< 527\)
\(\Rightarrow10x\in\left\{420;430;440;450;460;470;480;490;500;510;520\right\}\)
\(\Rightarrow x\in\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)
a: zz'\(\perp\)tt'
yy'\(\perp\)tt'
Do đó: zz'//yy'
=>\(\widehat{ABN}=\widehat{xAM}\)(hai góc đồng vị)
mà \(\widehat{xAM}=70^0\)
nên \(\widehat{ABN}=70^0\)
b:
\(\widehat{MAB}+\widehat{xAM}=180^0\)(hai góc kề bù)
=>\(\widehat{MAB}+70^0=180^0\)
=>\(\widehat{MAB}=110^0\)
yy'//zz'
=>\(\widehat{MAB}=\widehat{x'Bt'}\)(hai góc đồng vị)
=>\(\widehat{x'Bt'}=110^0\)
AC là phân giác của góc MAB
=>\(\widehat{MAC}=\widehat{BAC}=\dfrac{1}{2}\cdot\widehat{MAB}=55^0\)
Xét ΔABC có \(\widehat{ACN}\) là góc ngoài tại đỉnh C
nên \(\widehat{ACN}=\widehat{ABC}+\widehat{BAC}\)
\(=55^0+70^0=125^0\)
c: Bk là phân giác của \(\widehat{zBx'}\)
=>\(\widehat{x'Bk}=\dfrac{\widehat{x'Bt'}}{2}=\dfrac{110^0}{2}=55^0\)
=>\(\widehat{x'Bk}=\widehat{BAC}\)
mà hai góc này là hai góc ở vị trí đồng vị
nên Bk//AC