\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100^2}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{100^2}\)

\(B=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)

\(B=\frac{1\cdot101}{100\cdot2}=\frac{101}{200}\)

S = 1 - 1/2² - 1/3² - 1/4² -.. - 1/100² 
- - - 
Có: 1/k² < 1/(k-1)k = 1/(k-1) - 1/k (với mọi k nguyên, k > 1) 

1/2² < 1 /1.2 = 1/1 - 1/2 
1/3² < 1 /2.3 = 1/2 - 1/3 
... 
1/10² < 1 /9.100 = 1/9 - 1/100 
+ + + cộng vế lại + + + 
1/2² + 1/3² +..+ 1/10² < 1 - 1/100 

=> -1/2² - 1/3² - .. - 1/100² > -1 + 1/100 

=> 1 - 1/2² - 1/3² - .. - 1/100² > 1/100 > 0 (đpcm) 

♥Tomato♥

Hình đây nhé mn!!

B A C H M

Mn thg cảm, hình ko đc  chính xác cho lắm ^^"

tự ans lun vậy.--"

Từ H kẻ HE \(\perp AB,HF\perp AC\)

Xét \(\Delta HEB\)và \(\Delta AFM\)có:

\(\widehat{BEH}=\widehat{MFH}\left(=90^0\right)\)

HB=HM(gt)

\(\widehat{EBH}=\widehat{HMF}\)(cùng phụ với góc C)

\(\Rightarrow\Delta HEB=\Delta AFM\left(ch-gn\right)\)

=>HE=HF(2 CẠNH t/ư)

Xét \(\Delta HAE\)và \(\Delta HAF\)có:

\(\widehat{HEA}=\widehat{HFA}\left(=90^0\right)\)

HA chung

HE=HF(cmt)

\(\Rightarrow\Delta HAE=\Delta HAF\left(ch-cgv\right)\)

\(\Rightarrow\widehat{EAH}=\widehat{FAH}\)(2 góc t/ư)

\(\Rightarrow AH\)là pg \(\widehat{A}\)(đpcm)

Ta có :\(x-y⋮11\Rightarrow3x\left(x-y\right)⋮11\Rightarrow M⋮11\)

Ta có: \(x-y⋮11\Rightarrow\left(x-y\right)\left(x+y\right)⋮11\Rightarrow x\left(x+y\right)-y\left(x+y\right)⋮11\Rightarrow x^2+xy-xy-y^2⋮11\Rightarrow x^2-y^2⋮11\)

\(\Rightarrow-1\left(x^2-y^2\right)⋮11\Rightarrow y^2-x^2⋮11\Rightarrow N⋮11\)

Do đó\(\hept{\begin{cases}M⋮11\\N⋮11\end{cases}}\Rightarrow M-N⋮11\)(đpcm)

Vi x-y chia het cho 11 => 3x.(x-y) chia het cho 11=>M chia het cho 11 (1)

y^2-x^2=(y+x)(y-x).Vi x-y chi het cho 11 => y-x chia het cho11 =>(y+x)(y-x) chia het cho11<=> y^2-x^2 chia het cho 11

=> N chia het cho 11 (2)

Từ (1) và (2)=> M -N chia hết cho 11

=> Đpcm

IK ở đâu ra vậy

IK\(\perp\)BC (K thuộc BC)