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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)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(x^2+5x+5\right)^2\)
(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 )
\(x^4+2x^2-24\)
Đặt \(t=x^2\) ta có:
\(t^2+2t-24=t^2-4t+6t-24\)
\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t+6\right)\left(t-4\right)\)
\(=\left(x^2+6\right)\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2+6\right)\)
Ta có:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)
k nha!!
\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)
\(\text{Phân tích thành nhân tử :}\)
\(\left(x^2+10x+20\right)^2\)
(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+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
(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+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
a) a4 + a2 - 2
a4 + 2a2 - a2 - 2
a2.( a2 + 2 ) - ( a2 + 2 )
( a2 - 1 ).( a2 + 2 )
( a + 1 ).( a - 1 ).( a2 +2 )
b) x4 + 4x2 - 5
x4 + 5x2 - x2 - 5
x2.( x2 + 5 ) - ( x2 + 5 )
( x2 - 1 ).( x2 + 5 )
( x + 1 ).( x - 1 ).( x2 + 5 )
c) x3 - 19x - 30
x3 + 2x2 - 2x2 + 4x - 15x - 30
x2( x + 2 ) - 2x.( x + 2 ) - 15.( x + 2 )
( x + 2 ).( x2 - 2x - 15 )
d) x3 - 7x - 6
x3 - 3x2 + 3x2 - 9x + 2x - 6
x2.( x - 3 ) + 3x.( x - 3 ) + 2.( x - 3 )
( x - 3 ).( x2 + 3x +2 )
( x - 3 ).( x2 + 2x + x + 2 )
( x - 3 ).( x.( x + 2 ) + ( x + 2 )
( x + 1 ).( x + 2 ).( x - 3 )
e) x3 - 5x2 - 14x
x3 - 7x2 + 2x2 - 14x
x2.( x - 7 ) + 2x.( x - 7 )
( x - 7 ).( x2 + 2x )
x.( x + 2 ).( x - 7 )
Ta có:
x4+2x3+x2+x+1=(x2)2+2.x2.x+x2+x+1
=(x2+x)+(x+1)
=x2+2x+1
=(x+1)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