nà ní ko có quy luật à 

1/\(\frac{84^2-16^2}{37^2-63^2}=\frac{\left(84-16\right)\left(84+16\right)}{\left(37-63\right)\left(37+63\right)}=\frac{68.100}{-26.100}=\frac{-68}{26}=\frac{-34}{13}\)

2/ \(199^2=\left(200-1\right)^2=40000-400+1=39601\)

3/ \(31^2=\left(30+1\right)^2=900+60+1=961\)

4/ \(45.55=\left(50-5\right)\left(50+5\right)=50^2-25=2500-25=2475\)

5/ \(78.82=\left(80-2\right)\left(80+2\right)=80^2-4=6400-4=6396\)

phân tích nhân tử đi -_-, làm ra dài nên lười viết, chỉ cho cách làm đáy

ra đc ko?

\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(=\left(\frac{1}{200}-1\right)\left(\frac{1}{200}+1\right)\left(\frac{1}{199}-1\right)\left(\frac{1}{99}-1\right)...\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)

\(=\frac{-199}{200}.\frac{201}{200}.\frac{-198}{199}.\frac{200}{199}...\frac{-100}{101}.\frac{102}{101}\)

\(=\left(-\frac{199}{200}.\frac{-198}{199}...\frac{-100}{101}\right)\left(\frac{201}{200}.\frac{200}{199}...\frac{102}{101}\right)\)

\(=\frac{100}{200}.\frac{201}{101}=\frac{201}{202}\)

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x² + 5x - 2) - (x³ - 1) = 0

=> (x - 1)(x² + 5x - 2) - (x - 1)(x² + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50

b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)

\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)

f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)

\(\Leftrightarrow x=-66\)

a/ Tách 300 thành 100 chữ số 3 rồi chuyển vế dồn từng số 3 vào ( ) có \(\left(x^2-x-2\right)+\left(x^2-2x\right)+\left(x^2-3x+2\right)+...+\left(x^2-100x+196\right)\)

=0 \(\Leftrightarrow\left(x-2\right)\left(x+1\right)+x\left(x-3\right)+\left(x-1\right)\left(x-2\right)+...+\left(x-96\right)\left(x-4\right)+\left(x-97\right)\left(x-3\right)+\left(x-98\right)\left(x-2\right)\)=0\(\Leftrightarrow\left(x-2\right)\left(2x-97\right)+\left(x-3\right)\left(2x-97\right)+...=0\Rightarrow x=2\)

b tường đương \(x^2-4+\frac{4x^2}{x^2-4x+4}-1=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{3x^2+4x-4}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2+\frac{3x-2}{\left(x-2\right)^2}\right)=0\Leftrightarrow x=2\)

S=\(3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)

\(S=3\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\right)\)

\(S=3.\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)

\(S=\frac{3}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)

\(S=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(S=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(S=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)

S = 3 + 1 + 1/2 +....