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a) \(\frac{2x-1}{x^2-5x+6}\)
\(=\frac{5x-10-3x+9}{x^2-2x-3x+6}\)
\(=\frac{5\left(x-2\right)-3\left(x-3\right)}{x\left(x-2\right)-3\left(x-2\right)}\)
\(=\frac{5\left(x-2\right)-3\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{5\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)\(-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{5}{x-3}+\frac{-3}{x-2}\)
\(M=\frac{2x-1}{x^2-5x+6}=\frac{2x-1}{\left(x-2\right)\left(x-3\right)}=\frac{5\left(x-2\right)-3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{x-3}-\frac{3}{x-2}=\frac{5}{x-3}+\frac{3}{2-x}\)
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
\(\dfrac{x^3-x^2-x+1}{x^4-2x^2+1}=\dfrac{x^2\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{1}{x+1}\)
\(\dfrac{5x^3+10x^2+5x}{x^3+3x^2+3x+1}=\dfrac{5x\left(x+1\right)^2}{\left(x+1\right)^3}=\dfrac{5x}{x+1}\)
a) = \(\frac{2x}{\left(x-2\right)\left(x-3\right)}\)-\(\frac{1}{\left(x-2\right)\left(x-3\right)}\)
các bài sau tt
k hiểu