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LT
1
FT
1
6 tháng 12 2017
ta có B=2+\(\frac{x}{x-2}\)- \(\frac{4x^2}{x^2-4}\)- \(\frac{2-x}{x+2}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)+x\left(x+2\right)-4x^2-\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left\{2\left(x-2\right)+x\right\}-\left\{4x^2-\left(x-2\right)^2\right\}}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(3x-4\right)-\left(2x-x+2\right)\left(2x+x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left(3x-4\right)-\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2\right)\left(3x-4-3x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-2}{x-2}\)
28 tháng 7 2019
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
BH
28 tháng 7 2019
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
NN
22 tháng 10 2020
2. Đặt \(x-1996=t\)
\(\Rightarrow\left(x-1996\right)^3+\left(x-1997\right)^3-1=t^3+\left(t-1\right)^2-1\)
\(=t^3+t^2-2t+1-1=t^3+t^2-2t=t\left(t^2+t-2\right)\)
\(=t.\left[\left(t^2-t\right)+\left(2t-2\right)\right]=t\left[t\left(t-1\right)+2\left(t-1\right)\right]\)
\(=t\left(t-1\right)\left(t+2\right)=\left(x-1996\right)\left(x-1996-1\right)\left(x-1996+2\right)\)
\(=\left(x-1996\right)\left(x-1997\right)\left(x-1994\right)\)
22 tháng 10 2020
1. Đặt x2 + 4x + 8 = y
bthuc ⇔ y2 + 3xy + 2x2
= y2 + xy + 2xy + 2x2
= ( xy + y2 ) + ( 2x2 + 2xy )
= y( x + y ) + 2x( x + y )
= ( x + y )( y + 2x )
= ( x + x2 + 4x + 8 )( x2 + 4x + 8 + 2x )
= ( x2 + 5x + 8 )( x2 + 6x + 8 )
= ( x2 + 5x + 8 )( x2 + 2x + 4x + 8 )
= ( x2 + 5x + 8 )[ x( x + 2 ) + 4( x + 2 ) ]
= ( x2 + 5x + 8 )( x + 2 )( x + 4 )
2. Đặt t = x - 1996
bthuc ⇔ t3 + ( t - 1 )2 - 1
= t3 + t2 - 2t + 1 - 1
= t3 + t2 - 2t
= t( t2 + t - 2 )
= t( t2 - t + 2t - 2 )
= t( t - 1 )( t + 2 )
= ( x - 1996 )( x - 1996 - 1 )( x - 1996 + 2 )
= ( x - 1996 )( x - 1997 )( x - 1994 )
3. 4( x2 + 15x + 59 )( x2 + 18x + 72 ) - 3x2 < bó tay :)) >
\(\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)