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 )

\(x^4-x^3-2x-4\)

\(=x^4-x^3-2x^2+2x^2-2x-4\)

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

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

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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ĐỀ MÀ SAI THÌ LẤY CÁI CHI MÀ LÀM....???

#)Giải :

\(x^3-2x-4\)

\(=x^3+2x^2-2x^2+2x-4x-4\)

\(=x^3+2x^2+2x-2x^2-4x-4\)

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

\(=x^3-2x^2+2x^2-4x+2x-4\)

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

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

cảm ơn nha!

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

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

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)