x³ - 11x² + 30x = x(x² - 11x + 30) = x(x² - 6x - 5x + 30) = x[x(x - 6) - 5(x - 6)] = x(x - 5)(x - 6)

Trả lời:

x³ - 11x² + 30x

= x ( x² - 11x + 30 )

= x ( x² - 5x - 6x + 30 ) 

= x [ ( x² - 5x ) - ( 6x - 30 ) ] 

= x [ x ( x - 5 ) - 6 ( x - 5 ) ]

= x ( x - 6 ) ( x - 5 )

49 LÀ SỐ ĐÓ NHA

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Khó quá..............................................................................

Với \(x=0\Rightarrow y=\ne2\)

Với \(x>1\Rightarrow\)VT lẻ \(\Rightarrow y=2x+1\)

                   \(2^x+2=\left(2x+1\right)^2-1=4x\left(x+1\right)\)

 \(\Leftrightarrow2^{x-1}+1=2x\left(x+1\right)\)

do \(x>1\Rightarrow2^{x-1}\)chẵn \(\Rightarrow\)VT lẻ , mà VP chẵn

                                              \(\Rightarrow\)P/t vô nghiệm

Vậy p/t có nghiệm là \(\hept{\begin{cases}x=0\\y=\ne2\end{cases}}\)

45:8=5 (dư 5)

45 là đúng

45 nha

\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+4x^2+3}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+1\right)\left(x^2+3\right)}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^2}{x^4-x^2+1}\)

\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+1\right)\left(x^2+3\right)}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)

\(=\frac{x^4+2+\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(\frac{x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2}{x^4-x^2+1}\)

Vậy \(M=\frac{x^2}{x^4-x^2+1}\forall x\)

a.\(x^2+7x+6\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

Sửa đề:.\(x^4+2008x^2+2007x+2008\)

\(=x^4+x^2+1+2007x^2+2007x+2007\)

\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

Trả lời:

a, x² + 7x + 6

= x² + x + 6x + 6

= ( x² + x ) + ( 6x + 6 )

= x ( x + 1 ) + 6 ( x + 1 )

= ( x + 6 ) ( x + 1 )

\(x^3+x^2+a-x⋮\left(x+1\right)^2\)

\(\Leftrightarrow x^3+x^2+a-x=\left(x-1\right)\left(x+1\right)^2+a+1\)

Để \(x^3+x^2+a-x⋮\left(x+1\right)^2\)thì   \(a+1=0\) \(\forall a\)

\(\Rightarrow a=-1\)

x^3 + x^2 + a - x x^2 + 2x + 1 x - 1 x^3 + 2x^2 + x -x^2 + a - 2x -x^2 - 2x - 1 a - 1

Để \(x^3+x^2+a-x⋮\left(x+1\right)^2\Rightarrow a-1=0\Leftrightarrow a=1\)