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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)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5+y\right)\left(x+5-y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(x^2+2\right)\left(5x-7\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5-y\right)\left(x+5+y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(5x-7\right)\left(x^2+2\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
= (x^4-4x^3)+(3x^3-12x^2)+(2x^2-8x)-(2x-8)
= x^3.(x-4)+3x^2.(x-4)+2x.(x-4)-2.(x-4)
= (x-4).(x^3+3x^2+2x-2)
Tk mk nha
Đặt x2+10x+5=a.Ta có biểu thức là : a(a+8)+16=a2+8a+16=(a+4)2=(x2+10x+9)2
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
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)
a) \(x^2\)\(+\)\(6x\)\(+\)\(9\)
\(=\left(x+3\right)^2\)
b) \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)
\(=\left(x+1\right)^3\)
c) \(8x^3\)\(-\)\(\frac{1}{8}\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(10x\)\(-\)\(25\)\(-\)\(x^2\)
\(=\)\(-x^2\)\(+\)\(10\)\(-\)\(25\)
\(=-\left(x^2-10+25\right)\)
\(=-\left(x-5\right)^2\)
e) \(\frac{1}{25}x^2\)\(-\)\(64y^2\)
=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
(x2 + 6x + 5)(x2 - 10x + 21) - 20
= (x2 + x + 5x + 5)(x2 - 3x - 7x + 21) - 20
= (x + 1)(x + 5)(x - 3)(x - 7) - 20
= (x2 -2x - 3)(x2 - 2x- 35) - 20
Đặt x2 - 2x - 19 = a
=> (a + 16)(a - 16) - 20 = a2 - 256 - 20 = a2 - 276
= \(\left(a-2\sqrt{69}\right)\left(a+2\sqrt{69}\right)\)
= \(\left(a^2-2x-19-2\sqrt{69}\right)\left(x^2-2x-19+2\sqrt{69}\right)\)