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(x2 + x)2 - 4(x2 + x) - 12
= [(x2 + x)2 - 4(x2 + x) + 4] - 16
= (x2 + x - 2)2 - 16
= (x2 + x - 6)(x2 + x + 2)
= (x2 - 2x + 3x - 6)(x2 + x + 2)
= (x - 2)(x + 3)(x2 + x + 2)
Đặt t = x2 + x
bthuc ⇔ t2 - 4t - 12
= t2 - 6t + 2t - 12
= t( t - 6 ) + 2( t - 6 )
= ( t - 6 )( t + 2 )
= ( x2 + x - 6 )( x2 + x + 2 )
= ( x2 - 2x + 3x - 6 )( x2 + x + 2 )
= [ x( x - 2 ) + 3( x - 2 ) ]( x2 + x + 2 )
= ( x - 2 )( x + 3 )( x2 + x + 2 )
\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)
Ta có: \(x^3+9x^2+26x+24\)
\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)
\(=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+7x+12\right)\)
\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)
\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
a,\(8x^2-8xy+2x=2x\left(4x-8y+1\right)\)
b,\(\left(x^2+2x\right)\left(x^2+4x+3\right)-24=x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24=\left(t+1\right)\left(t-1\right)-24=t^2-5^2=\left(t+5\right)\left(t-5\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)( đặt t = x2 + 3x + 1 )
Ta có :
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+5=t\)
=> Đa thức trở thành
\(\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)
Thay vào ta được
Đt=\(\left(x^2+5x+5\right)^2\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\) (1)
Đặt \(x^2+5x+5=t\) thì (1)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(x^2+5x+5\right)^2\)
Ta có: (x+2)(x+4)(x+6)(x+8)+16
=[(x+2)(x+8)]+[(x+4)(x+6)]+16
\(=\left[x^2+10x+16\right]\left[x^2+10x+24\right]+16\) (1)
Đặt \(x^2+10x+16=t\), khi đó (1) trở thành:
\(t\left(t+8\right)+16=t^2+8t+16=\left(t+4\right)^2\)
Thay \(x^2+10x+16=t\), ta có: \(\left(x^2+10x+16+4\right)^2=\left(x^2+10x+20\right)^2\)
Có gì đó sai sai á nhờ :vv?
( x + 2 )( x + 4 )( x + 6 )( x + 8 ) + 16
= [ ( x + 2 )( x + 8 ) ][ ( x + 4 )( x + 6 ) ] + 16
= ( x2 + 10x + 16 )( x2 + 10x + 24 ) + 16 (*)
Đặt t = x2 + 10x + 20
(*) <=> ( t - 4 )( t + 4 ) + 16
= t2 - 16 + 16
= t2 = ( x2 + 10x + 20 )2