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1. Phân tích đa thức thành nhân tử
a, 1/4x^2-5xy+25y^2
b, (7x-4)^2-(2x+1)^2
c, (x-2)^2-4y
d, 125-x^6
a) \(\frac{1}{4}x^2-5xy+25y^2=\left(\frac{1}{2}x\right)^2-5xy+\left(5y\right)^2\)
\(=\left(\frac{1}{2}x-5y\right)^2\)
b) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4+2x+1\right)\times\left(7x-4-2x-1\right)=\left(9x-3\right)\times\left(5x-5\right)\)
\(=3\times5\times\left(3x-1\right)\times\left(x-1\right)=15\times\left(3x-1\right)\times\left(x-1\right)\)
c)\(\left(x-2\right)^2-4y^2=\left(x-2-2y\right)\left(x-2+2y\right)\)
d) \(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
a) \(x^2-5xy+6y^2\)
\(=x^2-3xy-2xy+6y^2\)
\(=x\left(x-3y\right)-2y\left(x-3y\right)\)
\(=\left(x-2y\right)\left(x-3y\right)\)
b) \(16\left(x-1\right)^2-36y^2\)
\(=\left(4x-4\right)^2-\left(6y\right)^2\)
\(=\left(4x+6y-4\right)\left(4x-6y-4\right)\)
c) \(4\left(x+y\right)-12\left(x+y\right)^2\)
\(=\left(x+y\right)\left[4-12\left(x+y\right)\right]\)
\(=4\left(x+y\right)\left[1-3x-3y\right]\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)
\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)
\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)
\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)
\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)
Đặt \(p=x^2-4,5x-8\)ta có :
\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)
\(A=p^2-\left(2,5x\right)^2+4x^2\)
\(A=p^2-6,25x^2+4x^2\)
\(A=p^2-2,25x^2\)
\(A=p^2-\left(1,5x\right)^2\)
\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)
Thay \(p=x^2-4,5x-8\)vào A ta có :
\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)
\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)
\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)
\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)
\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)
Đặt \(x^2-2x-8=t\)
Ta có : \(\left(t-5x\right)t+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)
\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)
\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)
Học tốt ~~
c) \(-\frac{x^4}{4}+2x^2y^3-4y^6=-\left(\frac{x^4}{4}-2x^2y^3+4y^6\right)=-\left[\left(\frac{x^2}{2}\right)^2-2.\frac{x^2}{2}.2y^3+\left(2y^3\right)^2\right]=-\left(\frac{x^2}{2}-2y^3\right)\)
b, x2-2x-y2+1
=(x-1)2-y2
(HĐT số 2)
=(x-y-1)(x+y-1)
(HĐT số 3)
c, (x2+x)2+4(x2+x)-12
=(x2+x)2+4(x2+x)+4-16
=(x2+x+2)2-16 (HĐT số 1)
=(x2+x+2-4).(x2+x+2+4)
(HĐT số 3)
=(x2+x-2).(x2+x+6)
=(x2+2x-x-2) .(x2+x+6)
=(x+2).(x-1). (x2+x+6)
a, x3+4x2-7x-10
=x3-2x2+6x2-12x+5x-10
=x2(x-2) + 6x(x-2)+5(x-2)
=(x-2)(x2+6x+5)
=(x-2)(x2+5x+x+5)
=(x-2)(x+5)(x+1)