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b) \(x^3-3x^2+2\)
\(=x^3-2x^2-x^2+2\)
\(=x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x^2-x-2\right)\left(x-2\right)\)
c) \(x^4y^4+64\)
\(=x^4y^4+16x^2+64-16x^2\)
\(=\left(x^2y^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2y^2-4x+8\right)\left(x^2y^2+4x+8\right)\)
d) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^6-x^4-x+x^3-1\right]\)
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
a) (x2-4x+3)(x2-10x+24)+8=((x2-x)-(3x-3))((x2-6x)-(4x-24))+8
=(x(x-1)-3(x-1))(x(x-6)-4(x-6))+8=(x-1)(x-3)(x-4)(x-6)+8=((x-1)(x-6))(x-3)(x-4))+8
=(x2-7x+6)(x2-7x+12)+8
Đặt x2-7x+6=a
Ta có : a(a+6)+8=a2+6a+8=(a+2)(a+4)=(x2-7x+8)(x2-7x+10)=(x2-7x+8)(x-5)(x-2)
b) Tương tự như câu a kết quả là (x-3)(x3+9x2+21x+9)
c) x4+x3+6x2+3x+9=(x4+x3+3x2)+(3x2+3x+9)=x2(x2+x+3)+3(x2+x+3)=(x2+x+3)(x2+2)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)
b)\(\left(x^2-x+2\right)^2+4x^2-4x-4=\left(x^2-x+2\right)^2+4\left(x^2-x-1\right)\)
Đặt \(k=x^2-x+2\) thì biểu thức có dạng
k2+4(k-3)=k2+4k-12=k2-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x2-x)(x2-x+8)=(x-1)x(x2-x+8)
c)làm tương tự câu a
x^3 - 4x^2 + 4x + 4x - 8
= (X^3 - 8) - (4x^2 - 4x - 4x)
= (x - 2)(x^2 + 2x + 4) - 4x( x - 2)
= (x - 2)(x^2 + 2x + 4 - 4x)
= (x - 2)(x^2 - 2x + 4)
b) 4x^2 - 25 - (2x - 5)(2x- 7)
= (2x - 5)(2x + 5) - (2x - 5)(2x - 7)
= (2x - 5)(2x + 5 - 2x + 7)
= 12(2x - 5)
c) x^3 + 27 + (x + 3)(x - 9)
= (x+3)(x^2-3x+9) + (x + 3)(x - 9)
= (x + 3) (x ^2 -3x + 9 + x - 9)
= (x + 3)(x^2 - 2x) = x(x - 2)(x + 3)
dễ mà bạn ơi