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\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
minh moi bn vao link nay dang ky roi tra loi minigame nha : https://alfazi.edu.vn/question/5b7768199c9d707fe5722878
a, x4 - 3x3 - x + 3
= (x4 - x) - (3x3 - 3)
= x(x3 - 1) - 3(x3 - 1)
= (x - 3)(x3 - 1)
b, x2 - x - 12
= x2 - x - 16 + 4
= (x2 - 16) - (x - 4)
= (x2 - 42) - (x - 4)
= (x + 4)(x - 4) - (x - 4)
= (x + 4 - 1)(x - 4)
= (x + 3)(x - 4)
c, x2 - 7x + 12
= x2 - 3x - 4x + 12
= (x2 - 3x) - (4x - 12)
= x(x - 3) - 4(x - 3)
= (x - 4)(x - 3)
d, x2 - 2x - 8
= x2 - 4x + 2x - 8
= (x2 - 4x) + (2x - 8)
= x(x - 4) + 2(x - 4)
= (x + 2)(x - 4)
5, x2 - 10x + 21
= x2 - 3x - 7x + 21
= (x2 - 3x) - (7x - 21)
= x(x - 3) - 7(x - 3)
= (x - 7)(x - 3)
f, x7 - x2 - 1
= t không bt
a) \(x^3-5x^2+8x-4=\left(x^3-x^2\right)-4\left(x^2-x\right)+4\left(x-1\right)=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\) chia hết cho 2x-3 => 7 chia hết cho 2x -3
=> 2x -3 thuộc U(7) ={-7;-1;1;7}
+2x-3 =-7 => x =-2
+2x-3 =-1 => x =1
+2x-3 =1 => x =2
+2x -3 =7 => x =5
a/ \(x^3-5x^2+8x-4\)
= \(\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
b/ \(x^3-x^2+x-1\)
= \(\left(x^3-x^2\right)+\left(x-1\right)\)
= \(x^2\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+1\right)\)
\(\left(a\right)x^4-2x^3+3x^2-2x+1\)
\(\text{phân tích đa thức thành nhân tử:}\)
b) c) (x2 + x)(x2 + x + 1) - 2
d) (x + 1)(x + 2)(x + 3)(x + 4) - 3
\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2+4-4x\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)
a/ x4 +5x3 +10x-4
=(x4- 4)+(5x3 + 10x)
=(x2+2) (x2-2) + 5x(x2 +2 )
=(x2+2)(x2 -2 +5x)
b/x5 - x4 +x3 -x2 +x-1
=x4(x-1)+x3(x-1)+(x-1)
=(x-1)(x4+x3+1)