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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-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
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: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
2x – 2y – x2 + 2xy – y2
(Có x2 ; 2xy ; y2 ta liên tưởng đến HĐT (1) hoặc (2))
= (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
(Có x – y là nhân tử chung)
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.
a. $x^2-2x+1=(x-1)^2$
b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$
c. $5x^2-10xy=5x(x-2y)$
d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$
$=(x-y)(x+y+1)$
a) Cách 1.
Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)
= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).
Cách 2.
Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)
= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).
b) Biến đổi được a 4 - 9 rt 3 + a 2 -9a = (a- 9)a( a 2 +1).
c) Biến đổi được 3 x 2 + 5y - 3xy + (-5x) = (x - y)(3x - 5).
d) Biến đổi được x 2 - (a + b)x + ab = (x- a)(x - b).
e) Ta có 4 x 2 - 4xy + y 2 – 9 t 2 = ( 2 x - y ) 2 - ( 3 t ) 2
= (2x - y - 3t )(2x - y + 31).
g) Ta có x 3 - 3 x 2 y + 3 xy 2 - y 3 - z 3
= ( x - y ) 3 - z 3 = (x - y - z)( x 2 + y 2 + z 2 - 2xy + xz - yz).
h) Ta có x 2 - y 2 + 8x + 6y+ 7 = ( x 2 +8x + 16) - ( y 2 - 6y+ 9)
= ( x + 4 ) 2 - ( y - 3 ) 2 =(x-y + 7)(x + y + l).
a: Ta có: \(x^2-xy-3x+3y\)
\(=x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x-3\right)\)
b: Ta có: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c: Ta có: \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
\(a,=x\left(x-3\right)\\ b,=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\\ c,=\left(3x+7\right)\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(3x+9\right)=3\left(x+3\right)\left(x-2\right)\)