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Đơn giản thôi :]>
Sau khi phân tích thì P(x) có dạng ( x2 + dx + 2 )( x2 + ax - 2 )
P(x) = x4 - x3 - 2x - 4 = ( x2 + dx + 2 )( x2 + ax - 2 )
⇔ x4 - x3 - 2x - 4 = x4 + ax3 - 2x2 + dx3 + adx2 - 2dx + 2x2 + 2ax - 4
⇔ x4 - x3 - 2x - 4 = x4 + ( a + d )x3 + adx2 + ( 2a - 2d )x - 4
Đồng nhất hệ số ta được :
\(\hept{\begin{cases}a+d=-1\\ad=0\\2a-2d=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-1\\d=0\end{cases}}\)
( x2 + dx + 2 )( x2 + ax - 2 )
= ( x2 + 2 )( x2 - x - 2 )
= ( x2 + 2 )( x2 - 2x + x - 2 )
= ( x2 + 2 )[ x( x - 2 ) + ( x - 2 ) ]
= ( x2 + 2 )( x - 2 )( x + 1 )
=> P(x) = x4 - x3 - 2x - 4 = ( x2 + 2 )( x - 2 )( x + 1 )
a)x3+3x2+3x+1
=x3+3x2*1+3x*12+13
=(x+1)3
b)(x+y)2-9x2
=y2+2xy+x2-9x2
=y2-2xy+4xy-8x2
=y(y-2x)+4x(y-2x)
=(y-2x)(y+4x)
x7+x6+x5-x6-x5-x4+x5+x4+x3-x3-x2-x1+x2+x1+1
= x5(x2+x+1) - x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1) +(x2+x+1)
=(x2+x+1)( x5-x4+x3-x+1)
────(♥)(♥)(♥)────(♥)(♥)(♥) __ ɪƒ ƴσυ’ʀє αʟσηє,
──(♥)██████(♥)(♥)██████(♥) ɪ’ʟʟ ɓє ƴσυʀ ѕɧα∂σѡ.
─(♥)████████(♥)████████(♥) ɪƒ ƴσυ ѡαηт тσ cʀƴ,
─(♥)██████████████████(♥) ɪ’ʟʟ ɓє ƴσυʀ ѕɧσυʟ∂єʀ.
──(♥)████████████████(♥) ɪƒ ƴσυ ѡαηт α ɧυɢ,
────(♥)████████████(♥) __ ɪ’ʟʟ ɓє ƴσυʀ ρɪʟʟσѡ.
──────(♥)████████(♥) ɪƒ ƴσυ ηєє∂ тσ ɓє ɧαρρƴ,
────────(♥)████(♥) __ ɪ’ʟʟ ɓє ƴσυʀ ѕɱɪʟє.
─────────(♥)██(♥) ɓυт αηƴтɪɱє ƴσυ ηєє∂ α ƒʀɪєη∂,
───────────(♥) __ ɪ’ʟʟ ʝυѕт ɓє ɱє.
\(x^2\left(x+1\right)^2+x^2+\left(x+1\right)^2\)
\(=\left(x^2+1\right)\left(x+1\right)^2+x^2+1-1\)
\(=\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]-1\)
\(=\left(x^2+1\right)\left(x^2+1+2x+1\right)-1\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2+1-1\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)
\(=\left(x^2+1+x\right)^2\)
= (x2 + 1)(x+1)2 + x2 + 1 -1 = (x2 + 1) [(x+1)2 + 1] - 1 = (x2 + 1) (x2 + 1 + 2x + 1) - 1 = (x2 + 1)2 + 2x(x2 + 1) + x2 + 1 - 1
= (x2 + 1)2 + 2x(x2 + 1) + x2 = (x2 + 1 + x)2
=X^7+x^6+x^5=x^4+x^3+x^2+1-x^6-x^5-x^4-x^3
=x^5(x^2=x+1)+(x^2+1)-x^4(x^^2-x+1)
=(x^2+x+1)(x^5+x^2-x^4)-(x-1)(x^2+x+1)
=(x^2+1+x)(x^5+x^2-X^4-x+1)
mik lm rồi nên chắc đúng
\(x^7+x^2+1=x^7+x^6+x^5-x^6-x^5-x^4+x^4+x^2+x+1-x\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
Đặt x^2 + x + x = t
Ta có BT : \(t\left(t+1\right)-1^2=t^2+t-1\):)) đề lỗi j ko ?