$\displaystyle \begin{array}{{>{\displaystyle}l}} b) \ ( 2x-1)^{2} =( 3-x)^{2}\\ \rightarrow ( 2x-1)^{2} -( 3-x)^{2} =0\\ \rightarrow ( 2x-1-3+x)( 2x-1+3-x) =0\\ \rightarrow ( 3x-4)( x+2) =0\ \\ \rightarrow \begin{cases} 3x-4=0 & \\ x+2=0 & \end{cases}\rightarrow \begin{cases} x=\frac{4}{3} & \\ x=-2 & \end{cases}\\ c) \ 121x^{4} -y^{2} =\left( 11x^{2} -y\right)\left( 11x^{2} +y\right)\\ d) 125x^{3} -\frac{y^{3}}{27} =\left( 5x-\frac{y}{3}\right)\left( 25x^{2} +\frac{5xy}{3} +\frac{y^{2}}{9}\right)\\ e) 27+x^{3} y^{6} =3^{3} +\left( xy^{2}\right)^{3}\\ =\left( 3+xy^{2}\right)\left( 9-3xy^{2} +x^{2} y^{4}\right)\\ Bài\ 2:\ \\ 100^{2} -99^{2} +98^{2} -97^{2} +...+2^{2} -1\\ =( 100-99)( 100+99) +( 98-97)( 98+97) +..+( 2-1)( 2+1)\\ =100+99+98+97+...+2+1\\ =\frac{( 100+1) .100}{2} =5050\\ b) 48\left( 5^{2} +1\right)\left( 5^{4} +1\right)\left( 5^{8} +1\right)\left( 5^{16} +1\right)\left( 5^{32} +1\right)\\ 2.( 25-1)\left( 5^{2} +1\right)\left( 5^{4} +1\right)\left( 5^{8} +1\right)\left( 5^{16} +1\right)\left( 5^{32} +1\right)\\ 2\left( 5^{4} -1\right)\left( 5^{4} +1\right)\left( 5^{8} +1\right)\left( 5^{16} +1\right)\left( 5^{32} +1\right)\\ 2\left( 5^{8} -1\right)\left( 5^{8} +1\right)\left( 5^{16} +1\right)\left( 5^{32} +1\right)\\ 2\left( 5^{16} -1\right)\left( 5^{16} +1\right)\left( 5^{32} +1\right)\\ 2\left( 5^{32} -1\right)\left( 5^{32} +1\right)\\ =2.\left( 5^{64} -1\right)\\ =2.5^{64} -2\\ bài\ 4:\ CMR:\ ( 4n+3)^{2} -25\\ =( 4n+3-5)( 4n+3+5)\\ =( 4n-2)( 4n+8)\\ 2( 2n-1) .4( n+2) \ \\ =8( 2n-1)( n+2) \ \\ \rightarrow ( 4n+3)^{2} -25\ chia\ hết\ cho\ 8 \end{array}$

$\\$

Bài `1.`

`b,`

`(2x-1)^2 = (3-x)^2`

TH1 :

`2x-1=3-x`

`->2x+x=3+1`

`->3x=4`

`->x=4/3`

TH1 :

`2x-1=-3+x`

`->2x-x=-3+1`

`->x=-2` 

Vậy pt có tập nghiệm : `S={4/3; -2}`

`c,`

`121x^4 -y^2`

`= (11x^2)^2 - y^2`

`= (11x^2 - y) (11x^2 + y)`

`d,`

`125x^3 - y^3/27`

`= (5x)^3 - (1/3 y)^3`

`= (5x - 1/3y) (25x^2 + 5/3xy + 1/9y)`

`e,`

`27 +x^3y^6`

`=3^3 + (xy^2)^3`

`= (3 +xy^2) (9 - 3xy^2 + x^2y^4)`

$\\$

Bài `2.`

`a,`

`100^2 - 99^2 + ... +2^2 - 1`

`= (100^2 - 99^2) + ... + (2^2 - 1^2)`

`= (100 - 99)(100 + 99) + ... + (2-1)(2+1)`

`= 199 + ... + 3`

Số các số hạng của tổng : `(199-3):4+1=50` (số hạng)

`->` Tổng :

`= ( (199+3) . 50)/2`

`= 5050`

`b,`

`48 (5^2  +1)(5^4 + 1)(5^8 +1)(5^{16}+1)(5^{32}+1)`

`= 2 . (5^2 - 1) (5^2  +1)(5^4 + 1)(5^8 +1)(5^{16}+1)(5^{32}+1)`

`= 2 (5^4 - 1)(5^4 +1) (5^8 +1)(5^{16}+1)(5^{32}+1)`

`=2 (5^8 - 1)(5^8 +1)(5^{16}+1)(5^{32}+1)`

`= 2 (5^{16}-1)(5^{16}+1) (5^{32}+1)`

`= 2(5^{32}-1)(5^{32}+1)`

`= 2 . (5^{64}-1)`

`= 2 . 5^{64}-2`

$\\$

Bài `4.`

`(4n+3)^2 -25`

`= (4n+3)^2 - 5^2`

`= (4n+3 - 5)(4n +3+5)`

`= (4n-2) (4n+8)`

`= 16n^2 + 32n - 8n - 16`

`= 8 (2n^2 + 3n - 2)`

Vì `8` chia hết cho `8`

Mà `n` là số nguyên

`-> 8 (2n^2 +3n-2)` chia hết cho `8` với mọi số nguyên `n`

`-> (4n+3)^2 - 25` chia hết cho `8` với mọi số nguyên `n`