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a) 1/2(x3+8)=1/2(x+2)(x2-2x+4)
b) x4(x-y)+2x3(x-y)=x3(x+2)(x-y)
c) x2-(y2-6y+9)=x2-(y-3)2=(x-y+3)(x+y-3)
d) xy(x3+y3)=xy(x+y)(x2-xy+y2)
e)3x2(x2-25y2)=3x2(x-5y)(x+5y)
f) 4x4+4x2y2+y4-4x2y2= (2x2+y2)2-(2xy)2=(2x2-2xy+y2)(2x2+2xy+y2)
a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)
b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)
c) \(x^2-y^2+6y-9=x^2-\left(y-3\right)^2=\left(x+y-3\right)\left(x-y+3\right)\)
d) \(x^4y+xy^4=xy\left(x^3+y^3\right)=xy\left(x+y\right)\left(x^2-xy+y^2\right)\)
e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).
f) \(4x^4+y^4=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2+2xy\right)\left(2x^2-y^2-2xy\right)\)
a,15x^2-60=15x^2-15*4=15(x^2-4)
b,=(x^2+2xy+y^2)-25=(x+y)^2-5^2=[(x-y)+5][(x-y)-5]
xog 2 phần
a, x2-7x-14y+2x
=x(x+2)-7(x-2y)
b, x3-4x2y+4xy2-25x
=x3-4x2y+4xy2-y3-25x+y3
=(x-y)3-25x+y3
a ) = x(x+2) - 7(x+2y)
b) = -4 xy ( x-y) + (x^3-25x) [ câu này mk , chaqcs là làm đúng đâu ]
Bài 2:
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)
\(=25x^2+10x+1-\left(2xy-3\right)^2\)
\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)
\(=25x^2+10x+1-4x^2y^2+12xy-9\)
\(=25x^2-4x^2y^2+10x+12xy-8\)
Bài 2:
\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)
\(=x^3-1=x^3-9x^2+2x+6\)
\(=x^3-9x^2+2x+6=x^3-1\)
\(=x^3-9x^2+2x+6+1=x^3-1+1\)
\(=x^3-9x^2+2x+7=x^3\)
\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)
\(=-9x^2+2x+7=0\)
\(\Rightarrow x=-\frac{7}{9};x=1\)
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
1.
\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)
\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)
\(=5x^2-3xy^2+4y\)
2.
a) \(27x^4-8x=x\left(27x^3-8\right)\)
\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(16x^2y-4xy^2-4x^3+x^2y\)
\(=4xy\left(4x-y\right)-x^2\left(4x-y\right)\)
\(=x\left(4x-y\right)\left(4y-x\right)\)
c) \(x^2-2x-5+2\sqrt{5}\)
\(=\left(x-1\right)^2-6+2\sqrt{5}\)
\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)
\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)
Bài 1:
\(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)
\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)
\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)
\(=5x^2-3xy^2+4y\)
Bài 2:
a) \(27x^4-8x\)
\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)
\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(16x^2y-4xy^2-4x^3+x^2y\)
\(=4y^2+x^2-\left(4x^2\right)^2\)
\(=x\left(-4x^2+xy+4y^2\right)\)