Phân tích đa thức thành nhân tử:
a) x4 + 4
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24
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a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
a: \(A=x^3y-12xy-x^2y\)
\(=xy\cdot x^2-xy\cdot12-xy\cdot x\)
\(=xy\left(x^2-x-12\right)\)
\(=xy\left(x^2-4x+3x-12\right)\)
\(=xy\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=xy\left(x-4\right)\left(x+3\right)\)
c: \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
=(x+1)(x+4)(x+2)(x+3)-120
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
d: \(D=x^5-x^4+x^2-1\)
\(=\left(x^5-x^4\right)+\left(x^2-1\right)\)
\(=x^4\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4+x+1\right)\)
a: \(x\left(x^3-4\right)+2x^3-4\)
\(=x^4-4x+2x^3-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)-2x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2x-2\right)\)
b: \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=-\left(x^3-1+4x^2-4x\right)\)
\(=-\left[\left(x-1\right)\left(x^2+x+1\right)+4x\left(x-1\right)\right]\)
\(=-\left(x-1\right)\left(x^2+5x+1\right)\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(t=x^2+5x+4\)
(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)
Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)
a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
a. `6x(x-2015)-x+2015=6x(x-2015)-(x-2015)=(x-2015)(6x-1)`
b. `x^4+4x^2+4=(x^2)^2+2.x^2 .2 +2^2=(x^2+2)^2`
a) \(6x\left(x-2015\right)-x+2015\)
\(=6x\left(x-2015\right)-\left(x-2015\right)\)
\(=\left(x-2015\right)\left(6x-1\right)\)
b) \(x^4+4x^2+4\)
\(=x^4+2\cdot x^2\cdot2+2^2\)
\(=\left(x^2+2\right)^2\)
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!
Trả lời:
A=(x-1)(x+2)(x-3)(x+4)-144
A= (x2-5x-14)(x2-5x-24)-144 (1)
đặt m=x2-5x-14
=> A= m.(m-10)-144
A=m2-10m-144
A= (m-18)(m+8)
thay m vào, ta có:
A= (x2-5x-32)(x2-5x-6)
A=(x2-5x-32)(x+1)(x-6)
a) x4 + 4
=x4+4x2+4-4x2
=(x2+2)2-4x2
=(x2-2x+2)(x2+2x+2)
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24
=[(x+2)(x+5)][(x+3)(x+4)]-24
=(x2+7x+10)(x2+7x+12)-24
=(x2+7x+10)[(x2+7x+10)+2]-24
=(x2+7x+10)2+2(x2+7x+10)-24
=(x2+7x+10)2+2(x2+7x+10)+1-25
=(x2+7x+10+1)2-25
=(x2+7x+11)2-25
=(x2+7x+11-5)(x2+7x+11+5)
=(x2+7x+6)(x2+7x+18)
=(x2+x+6x+6)(x2+7x+18)
=[x.(x+1)+6.(x+1)](x2+7x+18)
=(x+1)(x+6)(x2+7x+18)
lưu ý bài b có nhiều cách
****