một số đề toán 8- phân tích thành nhân tử

T

thaolovely1412

tìm x
a)[TEX]5x(x-1)=x-1 \Leftrightarrow (x-1)(5x-1)=0 \Leftrightarrow[/TEX] [TEX]\left[\begin{x-1=0}\\{5x-1 = 0} [/TEX] \Leftrightarrow [TEX]\left[\begin{x=1}\\{x = \frac{1}{5}} [/TEX]
b)[TEX]2(x+5)-x^2-5x=0 \Leftrightarrow (2-x)(x+5)=0 \Leftrightarrow[/TEX] [TEX]\left[\begin{2-x=0}\\{x+5 = 0} [/TEX] \Leftrightarrow [TEX]\left[\begin{x=2}\\{x = -5} [/TEX]
 
T

thaolovely1412

Phân tich thành nhân tử
b) [TEX]5x^2+5xy-x-y=5x(x+y)-(x+y)=(5x-1)(x+y)[/TEX]
d)[TEX]16x-5x^2-3=15x-5x^2+x-3=(x-3)-5x(x-3)=(1-5x)(x-3)[/TEX]
a)[TEX]5x^2-10xy+5y^2-20z^2[/TEX][TEX]=5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5(x-y+2z)(x-y-2z)[/TEX]
 
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B

boy_100

2.tìm x
a)$5x(x-1)=x-1$
\Leftrightarrow$ 5x =1 $
\Leftrightarrow $x =\frac{1}{5}$
b)$2(x+5)-x^2-5x=0$
\Leftrightarrow $2(x+5) =x^2 +5x $
\Leftrightarrow $ 2(x+5) = x (x+5)$
\Leftrightarrow $x=2$
 
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