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Phân tích các đa thức sau thành nhân tử:
a,x3+4x-5
b,x3-3x2+4
c,x3+2x2+3x+2
d,x2+2xy+y2+2x-2y-3
e,(x2+3x)2-2(x2+3x)-8
f,(x2+4x+10)2-7(x2+4x+11)+7

31 tháng 7 2021 lúc 16:33

a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)

b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)

c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=$\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)$$=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)$$=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)$$=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)$$=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)$$=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)$=$\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)$

f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-$\left[7\left(x^2+4x+11\right)-7\right]$$=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)$$=\left(x^2+4x+10\right)\left(x^2+4x+3\right)$

31 tháng 7 2021 lúc 23:10

a) Ta có: $x^3+4x-5$

$=x^3-x+5x-5$

$=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)$

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

b) Ta có: $x^3-3x^2+4$

$=x^3+x^2-4x^2+4$

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

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

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

c) Ta có: $x^3+2x^2+3x+2$

$=x^3+x^2+x^2+x+2x+2$

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

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

d) Ta có: $x^2+2xy+y^2+2x+2y-3$

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

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

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

$=\left(x+y+3\right)\left(x+y-1\right)$

31 tháng 7 2021 lúc 23:12

e) Ta có: $\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8$

$=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8$

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

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

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

f) Ta có: $\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7$

$=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7$

$=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)$

$=\left(x^2+4x+3\right)\left(x^2+4x+10\right)$

$=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)$

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