Đáp án:
Giải thích các bước giải:
\(1)\ x^{2}-1=(x-1)(x+1)\\ 2)x^{2}-4=(x-2)(x+2)\\ 6)\ x^{2}-y^{2}=(x-y)(x+y)\\ 7)\ 4x^{2}-9=(2x)^{2}-3^{2}=(2x-3)(2x+3)\\ 11)\ 4x^{2}-25y^{2}=(2x)^{2}-(5y)^{2}=(2x-5y)(2x+5y)\\ 12)\ 16x^{4}-y^{2}=(4x^{2})^{2}-y^{2}=(4x^{2}-y)(4x^{2}+y)\\ 16)\ 25x^{4}-9y^{2}=(5x^{2})^{2}-(3y)^{2}=(5x^{2}-3y)(5x^{2}+3y)\\ 17)\ x^{6}-y^{6}=(x^{3})^{2}-(y^{3})^{2}=(x^{3}-y^{3})(x^{3}+y^{3})=(x-y)(x^{2}+xy+y^{2})(x+y)(x^{2}-xy+y^{2})\\ 21)\ x^{3}+8=x^{3}+2^{3}=(x+2)(x^{2}-2x+4)\\ 22)\ x^{3}+27=x^{3}+3^{3}=(x+3)(x^{2}-3x+9)\\ 26)\ 8x^{3}+27y^{3}=(2x)^{3}+(3y)^{3}=(2x+3y)(4x^{2}-6xy+9y^{2})\\ 27)\ \dfrac{1}{8}x^{6}+y^{3}=\bigg(\dfrac{1}{2}x^{2}\bigg)^{3}+y^{3}=\bigg(\dfrac{1}{2}x^{2}+y\bigg)\bigg(\dfrac{1}{4}x^{4}-\dfrac{1}{2}x^{2}y+y^{2}\bigg)\\ 31)\ x^{3}-1=(x-1)(x^{2}+x+1)\\ 32)\ x^{3}-8=x^{3}-2^{3}=(x-2)(x^{2}+2x+4)\\ 36)\ x^{6}-1=(x^{3})^{2}-1=(x^{3}-1)(x^{3}+1)=(x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1)\\ 37)\ 8x^{3}-27y^{3}=(2x)^{3}-(3y)^{3}=(2x-3y)(4x^{2}-6xy+9y^{2})\)
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