Toán 8

K

kute2linh

Bài 1)
a) x3+8y3=x3+(2y)3=(x+2y)(x2−2xy+y2)x^3+8y^3= x^3+ (2y)^3= (x+2y)(x^2- 2xy+y^2)
b)3x2−9x=3x(x−3)3x^2- 9x= 3x(x-3)
c)3x2−3xy−5x+5y=3x(x−y)−5(x−y)=(x−y)(3x−5)3x^2- 3xy- 5x+5y= 3x(x-y) - 5(x-y)= (x-y)(3x-5)
d)x3+5x2−6x=x(x2+5x−6)=x(x−1)(x+6)x^3+ 5x^2- 6x= x(x^2+5x-6)= x(x-1)(x+6)
 
K

kute2linh

Bài 3)
a)$x #2$

b) $x#2$
A= x+1x−2=x2−2x+1x−2=(x+1)2x−2x+\frac{1}{x-2}= \frac{x^2- 2x+1}{x-2}=\frac{(x+1)^2}{x-2}

b) $x#2$
Để A= 2
<=> (x+1)2x−2=2\frac{(x+1)^2}{x-2}= 2

=> x2−2x+1=2x−4x^2- 2x+1= 2x- 4
<=> x2−4x+5=0x^2- 4x+5=0
<=> (x−1)(x+5)(x-1)(x+5)
<=> x=1x=1 hoặc x=−5x=-5
 
Q

quynhphamdq

CÂU 1:
[TEX]a) x^3+8y^3=x^3+(2y)^3=(x+2y)(x^2-2xy+4y^2)[/TEX]
[TEX] b) 3x^2-9x=3x(x-3)[/TEX]
[TEX]c) 3x^2-3xy-5x+5y=x(3x-5)-y(3x-5)=(x-y)(3x-5)[/TEX]
[TEX]d)x^3+5x^2-6x=x(x^2+5x-6)=x(x^2-x+6x-6)=x(x-1)(x+6)[/TEX]
Câu 2:
ta có:
[TEX](\frac{1}{x^2-4x}+\frac{2}{16-x^2}+\frac{1}{4x+16}):\frac{1}{4x}[/TEX]
[TEX]=(\frac{1}{x^2-4x}-\frac{2}{x^2-16}+\frac{1}{4x+16}):\frac{1}{4x}[/TEX]
[TEX]=(\frac{1}{x(x-4)}+\frac{2}{(x-4)(x+4)}+\frac{1}{4(x+4)}:\frac{1}{4x}[/TEX]
[TEX]=\frac{4(x+4)-4.2x+(x-4)x}{4x(x-4)(x+4)}.4x[/TEX]
[TEX]=\frac{4x+16-8x+x^2-4x}{4x(x-4)(x+4)}.4x[/TEX]
[TEX]=\frac{x^2-8x+16}{(x-4)(x+4)}[/TEX]
[TEX]=\frac{(x-4)^2}{(x-4)(x+4)}[/TEX]
[TEX]=\frac{x-4}{x+4}[/TEX]
CÂU 3:
[TEX]a) x\not= \2[/TEX]
[TEX]b) A=x+\frac{1}{x-2}=\frac{x^2-2x+1}{x-2}=\frac{(x-1)^2}{x-2}[/TEX]
[TEX]c) A=\frac{x^2-2x+1}{x-2}=2 [/TEX]
\Rightarrow [TEX]x^2-2x+1=2(x-2)[/TEX]
\Rightarrow[TEX]x^2-2x+1-2x+4=x^2-4x+5=(x-2)^2+1=0(vô lí)[/TEX]
\Rightarrow x ko có nghiệm.
Vậy phương trình vô nghiệm khi A=2.
 
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