Phân tích đa thức thành nhân tử
X2 +5xy +3x3
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\(x^2-11x+3\\ =\left(x^2-4x+4\right)-7x-1\\ =\left(x-2\right)^2-\left(\sqrt{7x+1}\right)^2\\ =\left(x-2-\sqrt{7x+1}\right)\left(x-2+\sqrt{7x+1}\right)\)
\(=x^2+x-6x+6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x+6\right)\)
\(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x-4\right)\left(x+1\right)\)
\(x^2-6x+7=x^2-6x+9-2\\ =\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\\ x^4+64=x^4+16x^2+64-16x^2\\ =\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\\ a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2\\ =\left(a^2+2b^2\right)^2-4a^2b^2\\ =\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)
= x2 -7x -x +7
= x. (x-7) - (x-7)
= (x-1)(x-7)
Chúc bạn học tốt nha!
\(x^2-4-3\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
\(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left[\left(x-y\right)^2-1^2\right]+\left(3x-3y-3\right)\)
\(=\left[\left(x-y\right)-1\right]\left[\left(x-y\right)+1\right]+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left[\left(x-y+1\right)+3\right]\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
\(x^2+4xy+4y^2-25\)
\(=\left(x^2+4xy+4y^2\right)-25\)
\(=\left(x+2y\right)^2-5^2\)
\(=\left(x+2y+5\right)\left(x+2y-5\right)\)
= x( 2x+5y+3x2)