Giải thích vì sao ạ ^^

S=1/5^2+1/7^2+...+1/103^2

=>\(S< \dfrac{1}{5^2}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{101\cdot103}\)

=>\(S< \dfrac{1}{2\cdot5}+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)

=>S<1/25+49/515<5/32

a: \(=\left(4+\dfrac{3}{4}+\dfrac{1}{8}+3+\dfrac{1}{12}\right)+\left(-0.37-1.28-2.5\right)=\dfrac{191}{24}-\dfrac{83}{20}=\dfrac{457}{120}\)

b: \(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

sao giong nhau the

a) \(A=sin^254^o.tan17^o.tan73^o+cos54^o.sin34^o\)

\(=sin^254^o.tan17^o.cot17^o+cos54^o.cos54^o\)

\(=sin^254^o+cos^254^o=1\)

b) \(B=\dfrac{cosa+sina}{sina-cosa}=\dfrac{\dfrac{cosa}{sina}+1}{1-\dfrac{cosa}{sina}}=\dfrac{cota+1}{1-cota}\)

        \(=\dfrac{1+\sqrt{3}}{1-\sqrt{3}}=\dfrac{\left(1+\sqrt{3}\right)^2}{1-3}=\dfrac{1+2\sqrt{3}+3}{-2}=-2-\sqrt{3}\)

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