Giải thích: `x^2-5x+1`

`=x^2-2. 5/2x+25/4-21/4`

`=(x-5/2)^2-21/4`

`=(x-5/2-\sqrt{21}/2)(x-5/2+\sqrt{21}/2)`

`=(x-[5+\sqrt{21}]/2)(x-[5-\sqrt{21}]/2)`

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

\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-9\)

\(=\left(x^2+2\right)^2-9\)

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

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

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

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

a)  \(x^4+4x^2-5=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9\)

\(=\left(x^2+2+3\right)\left(x^2+2-3\right)=\left(x^2+5\right)\left(x^2-1\right)\)

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

b)  \(x^6-x^4+2x^3+2x^2=x^4\left(x^2-1\right)+2x^2\left(x+1\right)\)

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

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

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

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

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

\(=2x^4+6x^3+9x^2+6x+2\)(bạn nhân phá ngoặc rồi thu gọn nhé)

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

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

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

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

\(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(ĐKXĐ:x^2+y^2\ne0\)

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mk chưa lên lp 8

`a)5(x-y)-y(x-y)`

`=(x-y)(5-y)`

`b)x^2-6x-y^2+9`

`=(x^2-6x+9)-y^2`

`=(x-3)^2-y^2`

`=(x-3-y)(x-3+y)`