Đáp án:
Giải thích các bước giải:
\(1)\ x^{2}-6x+9=x^{2}-2.x.3+3^{2}=(x-3)^{2}\\ 2)\ 25+10x+x^{2}=5^{2}+2.5.x+x^{2}=(5+x)^{2}\\ 3)\ \dfrac{1}{4}a^{2}+2ab^{2}+4b^{4}=\bigg(\dfrac{1}{2}a\bigg)^{2}+2.\dfrac{1}{2}a.2b^{2}+(2b)^{2}=\bigg(\dfrac{1}{2}a+2b\bigg)^{2}\\ 4)\ \dfrac{1}{9}-\dfrac{2}{3}y^{4}+y^{8}=\bigg(\dfrac{1}{3}\bigg)^{2}-2.\dfrac{1}{3}.y^{4}+(y^{4})^{2}=\bigg(\dfrac{1}{3}-y^{4}\bigg)^{2}\\ 5)\ x^{3}+8y^{3}=x^{3}+(2y)^{3}=(x+2y)(x^{2}+2xy+4y^{2})\\ 6)\ 8y^{3}-125=(2y)^{3}-5^{3}=(2y-5)(4y^{2}+10y+25)\\ 7)\ a^{6}-b^{3}=(a^{2})^{3}-b^{3}=(a^{2}-b)(a^{4}+a^{2}b+b^{2})\\ 8)\ x^{2}-10x+25=x^{2}-2.x.5+5^{2}=(x-5)^{2}\\ 9)\ 8x^{3}-\dfrac{1}{8}=(2x)^{3}-\bigg(\dfrac{1}{2}\bigg)^{3}=\bigg(2x-\dfrac{1}{2}\bigg)\bigg(4x^{2}+x+\dfrac{1}{4}\bigg)\\ 10)\ x^{2}+4xy+4y^{2}=x^{2}+2.x.2y+(2y)^{2}=(x+2y)^{2}\\ 8)\ (3x+2)^{2}-4=(3x+2)^{2}-2^{2}=(3x+2-2)(3x+2+2)=3x(3x+4)\\ 9)\ 4x^{2}-25y^{2}=(2x)^{2}-(5y)^{2}=(2x-5y)(2x+5y)\\ 10)\ 4x^{2}-49=(2x)^{2}-7^{2}=(2x-7)(2x+7)\\ 11)\ 8z^{3}+27=(2z)^{3}+3^{3}=(2z+3)(4z^{2}+6z+9)\\ 12)\ \dfrac{9}{25}x^{4}-\dfrac{1}{4}=\bigg(\dfrac{3}{5}x^{2}\bigg)^{2}-\bigg(\dfrac{1}{2}\bigg)^{2}=\bigg(\dfrac{3}{5}x^{2}-\dfrac{1}{2}\bigg)\bigg(\dfrac{3}{5}x^{2}+\dfrac{1}{2}\bigg)\\ 13)\ x^{32}-1\\=(x^{16}-1)(x^{16}+1)\\ =(x^{8}-1)(x^{8}+1)(x^{16}+1)\\ =(x^{4}-1)(x^{4}+1)(x^{8}+1)(x^{16}+1)\\ =(x^{2}-1)(x^{2}+1)(x^{4}+1)(x^{8}+1)(x^{16}+1)\\ =(x-1)(x+1)(x^{2}+1)(x^{4}+1)(x^{8}+1)(x^{16}+1)\\ 14)\ 4x^{2}+4x+1=(2x)^{2}+2.2x.1+1^{2}=(2x+1)^{2}\\ 15)\ x^{2}-20x+100=x^{2}-2.x.10+10^{2}=(x-10)^{2}\\ 16)\ (y^{4}-14y^{2}+49)=(y^{2})^{2}-2.y^{2}.7+7^{2}=(y^{2}-7)^{2}\\ 17)\ 125x^{3}-64y^{3}=(5x)^{3}-(4y)^{3}=(5x-4y)(25x^{2}+20xy+16y^{2})\)
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