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a) co sai de ko
b)x3-2x2+4x2-8x+3x-6=x2(x-2)+4x(x-2)+3(x-2)=(x-2)(x2+4x+3)=(x-2)(x+3)(x+1)
c)x3-2x2+2x2-4x-3x+6=x2(x-2)+2x(x-2)-3(x-2)=(x-2)(x2+2x-3)=(x-2)(x+3)(x-1)
d)x3-3x2+x2-3x-2x+6=x2(x-3)+x(x-3)-2(x-3)=(x-3)(x2+x-2)=(x-3)(x+2)(x-1)
\(a,x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(b,2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
a) \(x^3-2x^2+x+xy^2\)
\(=x\left(x^2-2x+1+y^2\right)\)
\(=x\left[\left(x-1\right)^2+y^2\right]\)
\(=-x\left[\left(x-1\right)^2-y^2\right]\)
\(=-x\left(x-1+y\right)\left(x-1-y\right)\)
b) \(4x^2+16x+16\)
\(=4\left(x^2+4x+4\right)\)
\(=4\left(x+2\right)^2\)
a, 4x2 - 12x + 9
= (2x + 3)2
b, 9x4y3 + 3x2y4
= 3x2y3(3x2 + y)
c, ( x - 3 )2 - 2x ( x - 3 )
= (x - 3)(x - 3 - 2x)
= (x - 3)(-x - 3)
d, 3x ( x - 1 ) + 6 ( x - 1 )
= 3(x - 1)(x + 2)
e, 2x ( x + 1 ) - 4x - 4
= 2x(x + 1) - 4(x + 1)
= (x + 1)(2x - 4)
= 2(x + 1)(x - 2)
f, ( 2x - 3 )2 - 4x + 6
= (2x - 3)2 - 2(2x - 3)
= (2x - 3)(2x - 3 - 2)
= (2x - 3)(2x - 5)
\(B=\left(x^2+2x\right)-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3\) \(\left(1\right)\)
Đặt \(x^2+2x=t\) , khi đó \(\left(1\right)\Leftrightarrow t^2-2t-3=\left(t+1\right)\left(t-3\right)=\left(x^2+2x+1\right)\left(x^2+2x-3\right)=\left(x+1\right)^2\left(x-1\right)\left(x+3\right)\)
a) 2x3 + 8x2 - 8x
= 2x(x2 + 4x - 4)
= 2x(x2 + 4x + 4 - 8)
= 2x[(x + 2)2 - 8]
= \(2x\left(x+2-\sqrt{8}\right)\left(x+2+\sqrt{8}\right)\)
b) a2 - b2 + 4a + 4b
= (a - b)(a + b) + 4(a + b)
= (a + b)(a - b + 4)
c) x2 - 2x - 3
= x2 + x - 3x - 3
= x(x + 1) - 3(x + 1)
= (x + 1)(x - 3)
d) x2 - 4x - 3
= x2 - 4x + 4 - 7
= (x + 2)2 - 7
= \(\left(x+2-\sqrt{7}\right)\left(x+2+\sqrt{7}\right)\)
a) \(ab-ac-b^2+2bc-c^2\)
\(=\left(ab-ac\right)-\left(b^2-2bc+c^2\right)\)
\(=a\left(b-c\right)-\left(b-c\right)^2\)
\(=\left(a-b+c\right)\left(b-c\right)\)
b) \(x^6+8=\left(x^2\right)^3+2^3\)
\(=\left(x^2+2\right)\left(x^4-2x^2+4\right)\)
c) \(64x^3-8=\left(4x\right)^3-2^3\)
\(=\left(4x-2\right)\left(16x^2+8x+4\right)\)
\(=8\left(2x-1\right)\left(4x^2+2x+1\right)\)
d) \(x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x^2+4\right)\left(x-2\right)\)
a) x3-4x2+x+6=x3-3x2-(x2-x-6) = x2(x-3) - (x-3)(x+2) = (x-3)(x2-x-2) = (x-3)(x-2)(x+1)
hình như câu b là x3 + 2x2 -x -2 đúng k. đặt nhân tử chung là ra thôi.
good luck
a) Nhẩm nghiệm nha. Ta được x=-1 thì x3 - 4x2 + x + 6=0 nên ta sẽ phân tích thành x+1 là nhân tử chung :
x3 - 4x2 + x + 6 = x3 +x2 - 5x2 - 5x + 6x + 6
= (x3 +x2 ) - (5x2 + 5x) + (6x +6)
= x2(x+1) - 5x(x+1) +6(x+1)
=(x+1)(x2 -5x+6)
=(x+1)(x2 - 2x - 3x + 6)
=(x+1)(x-2)(x-3)
Câu b bạn kiểm đề lại dùm mình nha