=(x-1)(3x+y)-(x-1)

=(x-1)(3x+y-1)

\(=x^2\left(y+1\right)-\left(y+1\right)\)

=(y+1)(x-1)(x+1)

ý a nhóm 2 cái đầu, 2 cái cuối

các ý còn lại lấy máy bấm nghiệm ra rồi phân tích

Bn ưi bạn có thể làm cụ thể cho mik đc ko!!! Bật mí nha: Mik ko bit bấm nghiệm hi hi, còn câu a mik cx nhóm như vậy mak nó hổng có ra chắc là do mik làmsai chỗ nào đó rùi, bạn có thể giải ra để mik bit mik sai chổ nào ko, cảm ơn bn nhìu nek!!!^-^!!!

a)  \(x^4-4x^2-6x+12=x^2\left(x^2-4\right)-6\left(x-2\right)\)

\(=x^2\left(x+2\right)\left(x-2\right)-6\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3+2x^2-6\right)\)

b)  \(x^2\left(x^2+4\right)-x^2+4=x^4+3x^2+4\)

\(=x^4+4x^2+4-x^2=\left(x^2+2\right)^2-x^2=\left(x^2-x+2\right)\left(x^2+x+2\right)\)

c)  \(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)

d)  \(x^2-7xy+12y^2=x^2-3xy-4xy+12y^2\)

\(=x\left(x-3y\right)+4y\left(x-3y\right)=\left(x-3y\right)\left(x+4y\right)\)

e)  \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\)

\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x+2\right)\)

a)\(\frac{1}{64}x^6-125y^3=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

b)\(x^6+1=\left(x^2\right)^3+1^3=\left(x^2+1\right)\left(x^4+x^2+1\right)\)

c)\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

d)\(x^9+1=\left(x^3\right)^3+1=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)

\(=x^3\left(x+1\right)\left(x^2-x+1\right)\left(x^2-x+1\right)\)

a. 12xy² - 8x²y = 4xy . (3y - 2x)

b. 3x + 3y - x² - xy = (3x + 3y) - (x² + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)

GIUSP MIK VS MN ƠI

Bài 2:

a: =>2x^2-8+3x-2x^2=15

=>3x=23

=>x=23/3

b: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

=>x=5 hoặc x=-3

c: =>x^2+7x-30=0

=>(x+10)(x-3)=0

=>x=3 hoặc x=-10

a(b³ - c³) + b(c³ - a³) + c(a³ - b³) 

= ab³ - ac³ + bc³ - ba³ + ca³ - cb³

= b(c³ - a³ - cb² + ab²) - ac(c² - a²) 

= b[(c - a)(a² + ac + c²) - b²(c - a)] - ac(c - a)(c + a) 

= (c - a)[a²b + abc + bc² - b³ - ac² - a²c]

= (c - a)[b(c² - b²) - ac(c - b) - a²(c -b)]

= (c - a)(c - b)[b(b + c) - ac - a²] = (c - a)(c - b)(b² + bc - ac - a²]

= (c - a)(c - b)[(b - a)(b + a) + c(b - a)] 

= (c - a)(c - b)(b - a)(a + b + c)

(a + b)³ - (a - b)³ 

 = (a + b - a + b)[(a + b)² + (a + b)(a - b) + (a - b)²]

= 2b(a² + 2ab + b² + a² - b² + a² - 2ab + b²]

= 2b(3a² + b²]

x³ - 3x² + 3x - 1 - y³

= (x - 1)³ - y³ 

 = (x - y - 1)(x² - 2x + 1 + xy - y + y²]

x^{m + 4} + x^{m + 3} - x - 1 

= x^{m + 3}(x +1) - (x + 1)

= (x^{m + 3} - 1)(x + 1) 

= (x - 1)[x^{m + 2} + x^{m + 1} + .... + 1](x + 1)

\(x^2+4xy+4y^2-4z^2-1-4z\)

\(=x^2+4xy+4y^2-\left(4z^2+4z+1\right)\)

\(=\left(x+2y\right)^2-\left(2z+1\right)^2\)

\(=\left(x+2y+2z+1\right)\left(x+2y-2z-1\right)\)

Ta có: \(x^5-2x^4+3x^3-4x^2+2\)

\(=x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)

\(=x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x^4-x^3+2x^2\right)-2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^4-x^3+2x^2-2x-2\right)\)