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a) \(x^2+2xy+x+2y\)
\(=x\left(x+2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+1\right)\)
b) \(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
c) \(x^2-6x+9-9y^2\)
\(=\left(x^2-6x+9\right)-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
d) \(x^3-3x^2+3x-1+2\left(x^2-x\right)\)
\(=\left(x^3-1\right)-\left(3x^2-3x\right)+2\left(x^2-x\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)+2x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-3x+2x\right)\)
\(=\left(x-1\right)\left(x^2+1\right)\)
e) \(15\left(x-y\right)-25x+25y\)
\(=15\left(x-y\right)-25\left(x-y\right)\)
\(=\left(15-25\right)\left(x-y\right)\)
\(=-10\left(x-y\right)\)
f) \(12x^2-3xy+8xz-2yz\)
\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x+2z\right)\)
y) \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
Phân tích đa thức thành nhân tử:
a) (x-1)(x-2)(x-3)(x-4)+1
b) (x2+3x+2)(x2+7x+12)+1
c) 12x2-3xy-8xz+2yz
a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)
\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)
\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)
Đặt \(a=x^2-5x+5\)
\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)
\(\Leftrightarrow A=a^2-1^2+1\)
\(\Leftrightarrow A=a^2\)
Thay \(a=x^2-5x+5\)vào A ta có :
\(A=\left(x^2-5x+5\right)^2\)
b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)
\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)
\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)
\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
Làm tương tự câu a)
c) \(12x^2-3xy-8xz+2yz\)
\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x-2z\right)\)
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a) \(x^2-3xy+x-3y=x\left(x-3y\right)+\left(x-3y\right)=\left(x-3y\right)\left(x+1\right)\)
b) \(x^2-6x-y^2+9=x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
c) \(7x^3y-14x^2y+7xy=7xy\left(x^2-2x+1\right)=7xy\left(x-1\right)^2\)
\(x^2-3xy+x-3y=\left(x^2+x\right)-\left(3xy+3y\right)=x\left(x+1\right)-3y\left(x+1\right)=\left(x+1\right)\left(x-3y\right)\)
\(x^2-6x-y^2+9=\left(x^2-2.x.3+3^2\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
\(7x^3y-14x^2y+7xy=\left(7x^3y-7x^2y\right)-\left(7x^2y-7xy\right)=7x^2y.\left(x-1\right)-7xy.\left(x-1\right)\)
\(=\left(x-1\right).\left(7x^2y-7xy\right)=7xy.\left(x-1\right).\left(x-1\right)=7xy.\left(x-1\right)^2\)
a,\(x^2y^2+y^3+zx^2+yz=\left(x^2y^2+y^3\right)+\left(zx^2+yz\right)\)
\(=y^2\left(x^2+y\right)+z\left(x^2+y\right)\)
\(=\left(y^2+z\right)\left(x^2+y\right)\)
b,\(x^4+2x^3-4x-4=x^4+2x^3+x^2-x^2-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
c,\(x^3+2x^2y-x-2y=\left(x^3+2x^2y\right)-\left(x+2y\right)\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2y\right)\)
7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, ĐK x >= 0
\(x-2\sqrt{x}-3=x-3\sqrt{x}+\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(-4x^2-4x+10=-\left(4x^2+4x+1\right)+11\)
\(=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11;12 xem lại đề
13, \(-x^3+6xy^2-12xy^2+8y^3=-\left(x^3-6xy^2+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
Trả lời:
7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, \(x-2\sqrt{x}-3\left(ĐK:x\ge0\right)\)
\(=x-3\sqrt{x}+\sqrt{x}-3=\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(10-4x-4x^2=-\left(4x^2+4x-10\right)=-\left(4x^2+4x+1-11\right)=-\left[\left(2x+1\right)^2-11\right]\)
\(=-\left(2x+1\right)^2+11=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11,sửa đề: \(15x\left(x-3y\right)+20y\left(3y-x\right)=15x\left(x-3y\right)-20y\left(x-3y\right)=5\left(x-3y\right)\left(3x-4y\right)\)
12, \(25x^2-2=\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)\)
13, sửa đề: \(-x^3+6x^2y-12xy^2+8y^3=-\left(x^3-6x^2y+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
a) \(x^3-2x^2+2x-1^3\)
\(=x\left(x^2-2x+1\right)+x-1\)
\(=x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\)
b) \(x^2y+xy+x+1\)
\(=xy\left(x+1\right)+\left(x+1\right)\)
\(=\left(xy+1\right)\left(x+1\right)\)
c) \(ax+by+ay+bx\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(a+b\right)\left(x+y\right)\)
d) \(x^2-\left(a+b\right)x+ab\)
\(=x^2-ax-bx+ab\)
\(=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
e) Ko biết làm
f) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
A) 7X2 - 7Y2 - 14X + 14Y
= ( 7X2 - 7Y2 ) - ( 14X - 14Y )
= 7( X2 - Y2 ) - 14( X - Y )
= 7( X - Y )( X + Y ) - 14( X - Y )
= 7( X - Y )( X + Y - 14 )
B) X2 - Y2 + 14X + 49
= ( X2 + 14X + 49 ) - Y2
= ( X + 7 )2 - Y2
= ( X - Y + 7 )( X + Y + 7 )
C) X2 - Y2 - X + Y
= ( X2 - Y2 ) - ( X - Y )
= ( X - Y )( X + Y ) - ( X - Y )
= ( X - Y )( X + Y - 1 )
D) X2 + 12Y - Y2 - 36
= X2 - ( Y2 - 12Y + 36 )
= X2 - ( Y - 6 )2
= ( X - Y + 6 )( X + Y - 6 )
E) X3 + X2 - 9X - 9
= ( X3 + X2 ) - ( 9X + 9 )
= X2( X + 1 ) - 9( X + 1 )
= ( X + 1 )( X2 - 9 )
= ( X + 1 )( X - 3 )( X + 3 )
a) Ta có: \(27m\left(m+n\right)-m-n\)
\(=27m\left(m+n\right)-\left(m+n\right)\)
\(=\left(m+n\right)\left(27m-1\right)\)
b) Ta có: \(15x\left(x-y\right)-25x+25y\)
\(=15x\left(x-y\right)-25\left(x-y\right)\)
\(=5\left(x-y\right)\left(3x-5\right)\)
c) Ta có: \(12x^2-3xy+8xz-2yz\)
\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x+2z\right)\)
d) Ta có: \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)