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b,(4x2 - 25)-(2x-5)(2x+7)
=(2x)2-52 -(2x-5)(2x+7)
=(2x-5)(2x+5)-(2x-5)(2x+7)
=(2x-5)(2x+5-2x-7)
=(2x-5).(-2)
e,x2-4x-21
=x2-7x+3x-21
=x(x-7)+3(x-7)
=(x-7)(x+3)
f,x2-7x+12
= x2 -4x-3x+12
=x(x-4)-3(x-4)
(x-4)(x-3)
Bạn tham khảo nhé mk chỉ giúp được ngần đây thui
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) Ta có: \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)(1)
Đặt \(a=x^2+x\)
(1)\(=a^2-14a+24\)
\(=a^2-12a-2a+24\)
\(=a\left(a-12\right)-2\left(a-12\right)\)
\(=\left(a-12\right)\left(a-2\right)\)
\(=\left(x^2+x-12\right)\left(x^2+x-2\right)\)
\(=\left(x^2+4x-3x-12\right)\left(x^2+2x-x-2\right)\)
\(=\left[x\left(x+4\right)-3\left(x+4\right)\right]\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x+4\right)\left(x-3\right)\left(x+2\right)\left(x-1\right)\)
b) Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+2x-x-2\right)\)
\(=\left(x^2+x+6\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c) Ta có: \(x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
d) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(2)
Đặt \(x^2+5x=b\)
(2)\(=\left(b+4\right)\left(b+6\right)+1\)
\(=b^2+10b+24+1\)
\(=b^2+10b+25\)
\(=\left(b+5\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
e) Ta có: \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)(3)
Đặt \(c=x^2+8x\)
(3)\(=\left(c+7\right)\left(c+15\right)+15\)
\(=c^2+22c+105+15\)
\(=c^2+22c+120\)
\(=c^2+12c+10c+120\)
\(=c\left(c+12\right)+10\left(c+12\right)\)
\(=\left(c+12\right)\left(c+10\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)
\(=\left[x\left(x+6\right)+2\left(x+6\right)\right]\left(x^2+8x+10\right)\)
\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)
\(1,x^2-xy-2x+2y\)
\(x\left(x-2\right)-y\left(x-2\right)\)
\(\left(x-2\right)\left(x-y\right)\)
\(2,x^2+4x+4-y^2\)
\(\left(x+2\right)^2-y^2\)
\(\left(x+2-y\right)\left(x+2+y\right)\)
\(3,x^2+x+y-y^2\)
\(\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(\left(x+y\right)\left(x-y+1\right)\)
\(4,x^3-x^2-4x+4\)
\(x^2\left(x-1\right)-4\left(x-1\right)\)
\(\left(x-1\right)\left(x^2-4\right)\)
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)\)
\(1,\)
\(\left(x^2-x\right)^2+4\left(x^2+x\right)-12\)
\(=x^4-2x^3+x^2+4x^2+4x-12\)
\(=x^4-2x^3+5x^2+4x-12\)
help meeeeeeeeeeeee