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a) x^2 - 4 + ( x - 2 )^2
= ( x- 2 )(x + 2 ) + ( x- 2)^2
= ( x - 2 ) ( x + 2 + x - 2 )
= 2x (x-2)
b) x^3 - 2x^2 + x - xy^2
= x ( x^2 - 2x + 1 - y^2)
= x [ ( x - 1 )^2 - y^2 ]
= x(x - 1 - y)( x - 1 + y )
c) x^3 - 4x^2 - 12x + 27
= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27
= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )
Để hai lần nha
= ( x+ 3 )(x^2 - 7x + 9 )
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
hk tốt
^^
Đây là cách hiện đại :
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-\left(2x^3-2x\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x^2+1\right)-2x\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)-2x\right)\)
a,=\(x^4-x^3-x^3+x^2-x^2+x+x-1\)
cu hai so nhom 1 nhom roi dat thua so chung la xong
b,x^4+x^3+x^3+x^2+x^2+x+x+1
cu hai so lai nhom 1 nhom va dat thua so chung
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
b, \(x^3+2x^2+2x+1=\left(x^2+x+1\right)\left(x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x^2-x+9\right)\left(x-3\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, sai đề
a, \(\left(ab-1\right)^2+\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x-3\right)\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, cho mình sửa đề xíu
\(x^4+2x^3+2x^2+2x+1=\left(x+1\right)^2\left(x^2+1\right)\)
\(a)\) \(x^2-2x-4y^2-4y\)
\(=\)\(\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\)\(\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\)\(\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\)\(\left(x-2y-2\right)\left(x+2y\right)\)
\(=\)\(2\left(x-y\right)\left(x+2y\right)\)
Chúc bạn học tốt ~
a) Ta có x2 - 2x - 4y2 - 4y
= x2 - 2x + 1 - 4y2 - 4y - 1
= (x - 1)2 - (4y2 + 4y + 1)
= (x - 1)2 - (2y + 1)2
= (x - 1 - 2y - 1)(x - 1 + 2y + 1)
= (x - 2y - 1)(x + 2y)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
a)x4-4(x2+5)-25=x4-4x2-45=(x4-9x2)+(5x2-45)=x2(x2-9)+5(x2-9)=(x2-9)(x2+5)=(x-3)(x+3)(x2+5)
b)a2-b2-2a+1=(a2-2a+1)-b2=(a-1)2-b2=(a-b-1)(a+b-1)
c)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
d)x2+4x-y2+4=(x2+4x+4)-y2=(x+2)2-y2=(x-y+2)(x+y+2)
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right).\left(x+2y\right)-2.\left(x+2y\right)\)
\(=\left(x+2y\right).\left(x-2y-2\right)\)
b) \(x^4+2x^3-4x-4=\left(x^4-4\right)+\left(2x^3-4x\right)=\left(x^2+2\right).\left(x^2-2\right)+2x.\left(x^2-2\right)\)
\(=\left(x^2-2\right).\left(x^2+2+2x\right)\)
c) \(x^2.\left(1-x\right)^2-4x-4x^2=x^2.\left(x^2-2x+1\right)-4x-4x^2=x^4-2x^3+x^2-4x-4x^2\)
\(x^4-2x^3-3x^2-4x=x.\left(x^3-2x^2-3x-4\right)\)
d) \(\left(1+2x\right).\left(1-2x\right)-x.\left(x+2\right).\left(x-2\right)=1-4x^2-x.\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x=1-x^3+4x-4x^2=\left(1-x\right).\left(1+x+x^2\right)+4x.\left(1-x\right)\)
\(=\left(1-x\right).\left(1+x+x^2+4x\right)=\left(1-x\right).\left(x^2+5x+1\right)\)
e) \(x^2+y^2-x^2y^2+xy-x-y=\left(x^2-x\right)-\left(x^2y^2-y^2\right)+\left(xy-y\right)\)
\(=x.\left(x-1\right)-y^2.\left(x^2-1\right)+y.\left(x-1\right)=x.\left(x-1\right)-y^2.\left(x-1\right)\left(x+1\right)+y.\left(x-1\right)\)
\(=\left(x-1\right).\left(x-y^2.\left(x+1\right)+y\right)=\left(x-1\right).\left(x-xy^2-y^2+y\right)\)
\(=\left(x-1\right)\left[-\left(xy^2-x\right)-\left(y^2-y\right)\right]=\left(x-1\right)\left[-x\left(y^2-1\right)-y\left(y-1\right)\right]\)
\(=\left(x-1\right)\left[-x\left(y-1\right)\left(y+1\right)-y\left(y-1\right)\right]=\left(x-1\right)\left(y-1\right)\left(-x.\left(y+1\right)-y\right)\)
\(=\left(x-1\right)\left(y-1\right)\left(-xy-x-y\right)=-\left(x-1\right)\left(y-1\right)\left(xy+x+y\right)\)
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)
a, \(x^3-4x^2+12x-27\) \(=x^3-3x^2-x^2+3x+9x-27\)
= \(x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\) \(=\left(x-3\right)\left(x^2-x+9\right)\)
b, \(x^3+2x^2+2x+1\) \(=x^3+x^2+x^2+x+x+1\)
= \(x^2\left(x+1\right)+x\left(x+1\right)+x+1=\) \(\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^4-2x^3+2x-1=\) \(x^4-x^3-x^3+x^2-x^2+x+x-1\)
= \(x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+x-1\)
= \(\left(x-1\right)\left(x^3-x^2-x+1\right)\)
d, \(x^4+2x^3+2x^2+2x+1=\) \(x^4+x^3+x^3+x^2+x^2+x+x+1\)
= \(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
= \(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
Ta có : x3 - 4x2 + 12x - 27
= (x3 - 27) - (4x2 - 12x)
= (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9 - 4x)
= (x - 3)(x2 - x + 9)
b) https://olm.vn/hoi-dap/question/1004349.html tôi tự coppy tôi
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)
d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
a) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)