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\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
a) \(x^2-2x-15\)
\(\Leftrightarrow x^2-2x+1-16\)
\(\Leftrightarrow\left(x-1\right)^2-4^2\)
\(\Leftrightarrow\left(x-5\right)\left(x-3\right)\)
\(a,x^2-2x-15=\left(x^2-2x+1\right)-16.\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-5\right)\left(x+3\right)\)
\(a,\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)
\(b,27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
\(c,x^6-x^6\)
\(=0\)
\(d,10x-25-x^2\)
\(=-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
a) \(x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(=\left(x+3\right)\left(x+3\right)\)
b) \(10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(=-\left(x-5\right)\left(x-5\right)\)
c) \(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(\frac{1}{25}x^2-64y^2\)
\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)
b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)
c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)
d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)
a, = (x+3y)^2
b, = (x-1/2)(x+1/2)
c, = (x-5)^2
d, = (2x+3y)(4x^2-6xy+9y^2)
e, = (x^3-y)^2
f,= (x+3y)^3
=x3(x+2)-13x2+12x-26x+24
=x3(x+2)-x(13x-12)-2(13x-12)
=x3(x+2)-(13x-12)(x+2)
=(x+2)(x3-x-12x+12)
(x+2)[(x2-1)-12(x-1)]
=(x+2)[x(x-1)(x+1)-12(x-1)]
=(x+2)(x-1)[x(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x(x-3)+4(x+3)]
=(x+2)(x-1)(x-3)(x+4)
trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
A . 5(x-y)-y(x-y)
=(x6-y)(5-y)
B . x^2 - xy - 8x+8y
=(x^2-xy)-(8x-8y))
=x(x-y) - 8(x-y)
C. x^2-10x+25 - y^2
=(x^2 - 10x + 25 ) - y^2
=(x-5)^2 - y^2
=(x-5+y)(x-5-y)
D . x^3 - 3x^2-4x+12
=(x^3 - 3x^2 ) - (4x - 12)
=x^2 (x-3)-4(x-3)
=(x^2-4)(x-3)
=(x+2)(x-2)(x-3)
D . 2x^2-2y^2- 6x-6y
=(2^x - 2y^2) - (6x+ 6y)
=2(x^2 - y^2) - 6(x+y)
=2(x+y)(x-y) - 6(x+y)
=2(x+y)(x-y-3)
E . x^3 - 3x^2 + 3x - 1
=(x-1)^3
D.x^2+3x+2
=x^2+2x+x+2
=(x^2+2x)+(x+2)
=x(x+2)+(x+2)
=(x+2)(x+1)
\(1,9x^3-3x^2+3x-1\)
\(=3x^2.\left(3x-1\right)+\left(3x-1\right)\)
\(=\left(3x^2+1\right).\left(3x-1\right)\)
\(4,x^4-x^3-10x^2+2x+4\)
\(=x^4-3x^3-2x^2+2x^3-6x^2-4x-2x^2-6x-4\)
\(=x^2.\left(x^2-3x-2\right)+2x.\left(x^2-3x-2\right)-2.\left(x^2-3x-2\right)\)
\(=\left(x^2+2x-2\right).\left(x^2-3x-2\right)\)
\(x^4+2x^3+10x-25\)
\(=x^4+5x^2+2x^3+10x-5x^2-25\)
\(=\left(x^2+5\right)\left(x^2+2x-5\right)\)