2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

=(x-1)(x+1)(x-2)(x+2)

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)

a)  \(A=x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left[x\left(x+10\right)\right].\left[\left(x+4\right)\left(x+6\right)\right]+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

đặt     \(x^2+10x+12=t\)khi đó:

\(A=\left(t-12\right)\left(t+12\right)+128\)

\(=t^2-16=\left(t-4\right)\left(t+4\right)\)

bạn thay trở lại nhé

b)  \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

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

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

d)  \(x^7+x^2+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

e)  \(4x^4+81=4x^4+36x^2+81-36x^2=\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)

\(x^5+x^4+1\)

\(=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

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

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

g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

                                       =\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)

đặt \(x^2+7x+10=a\)

ta có \(a\left(a+2\right)-24=a^2+2a-24\)

                                      \(=a^2+2a+1-25\)

                                      \(=\left(a+1\right)^2-5^2\)

                                      \(=\left(a+1-5\right)\left(a+1+5\right)\)

                                     \(=\left(a-4\right)\left(a+6\right)\)

\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

a) = (x +5)² - 2² = (x+5 -2)(x+5 +2) = (x+3)(x+7)

b) = x(x² -1) -6(x-1)= x(x+1)(x-1) -6(x-1) = (x-1)(x(x+1)-6)

a) \(x^2+10x+21=x^2+3x+7x+21\)

                                \(=x\left(x+3\right)+7\left(x+3\right)=\left(x+7\right)\left(x+3\right)\)

từ ý b đến ý e dùng chức lăng EQN của máy tính

bạn mở mt bấm mode rồi ấn 5 rồi tìm cái có ax³+bx²+cx+d

b) nhập 1 = ; 0 =;-7=;6= rồi ấn = sao cho hết nghiệm

có \(X_1=2;X_2=1;X_3=-3\)

a,2x²-7x+6=(2x²-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x²+x-6=(x²+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x³+3x²+6x+4=x³+x²+2x²+2x+4x+4
=(x+1)(x²+2x+4)
d,x¹⁰+x⁵+1=(x¹⁰-x)+(x⁵-x²)+(x²+x+1)
=x((x³)³-1)+x²(x³-1)+(x²+x+1)
=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+(x²+x+1)
=x(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+(x²+x+1)
(x²+x+1)(x²-x+x³-x²+1)
e,(12x²-12xy+3y²)-10x(2x-y)=3(4x²-4xy+y²)-10x(2x-y)
=3(2x-y)²-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)

phân tích đa thức thành nhân tử

a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)

\(2x^2-7x+6\)

\(=2x^2-3x-4x+6\)

\(=x\left(2x-3\right)-2\left(2x-3\right)\)

\(=\left(x-2\right)\left(2x-3\right)\)

      \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2\left(x^2+3x\right)+1\)

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

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

Chúc bạn học tốt.

\(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

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

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

https://olm.vn/hoi-dap/tim-kiem?q=Cho+%CE%94ABC+vu%C3%B4ng+t%E1%BA%A1i+A+c%C3%B3+AB+%3E+AC,+M+l%C3%A0+%C4%91i%E1%BB%83m+tu%E1%BB%B3+%C3%BD+tr%C3%AAn+BC.+Qua+M+k%E1%BA%BB+Mx+vu%C3%B4ng+g%C3%B3c+v%E1%BB%9Bi+BC+v%C3%A0+c%E1%BA%AFt+AB+t%E1%BA%A1i+I+c%E1%BA%AFt+CA+t%E1%BA%A1i+D.a.+Ch%E1%BB%A9ng+minh+%CE%94ABC+%C4%91%E1%BB%93ng+d%E1%BA%A1ng+v%E1%BB%9Bi+%CE%94MDCb.+Ch%E1%BB%A9ng+minh:+BI.BA+=+BM.BCc.+Cho+g%C3%B3c+ACB+=+60o+v%C3%A0+S%CE%94CDB+=+60+cm2.+T%C3%ADnh+S%CE%94CMAGi%C3%BAp+m%C3%ACnh+c%C3%A2u+c+v%E1%BB%9Bi&id=573451

a) \(x^{12}-3x^6+1\)

\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)

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

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

b) \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)

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

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

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

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

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

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