ohaan tích đa thức sau thành nhân tữ 4x^4+36x^2+81-36x^2
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\(=4x^2-28x-8x+56=4x\left(x-7\right)-8\left(x-7\right)=4\left(x-2\right)\left(x-7\right)\)
\(81\left(y-4\right)^2-9x^2-36x-36=9\left\{9\left(y-4\right)^2-\left(x+4\right)^2\right\}=9\left\{3\left(y-4\right)-\left(x+4\right)\right\}\left[3\left(y-4\right)+\left(x+4\right)\right]\)
\(=9\left(3y+12-x-4\right)\left(3y+12+x+4\right)=9\left(3y-x+8\right)\left(3y+x+16\right)\)
4x2 - 36x + 56
= 4(x2 - 9x + 14)
= 4(x2 - 2x - 7x + 14)
= 4[x(x - 2) - 7(x - 2)]
= 4(x - 2)(x - 7)
\(4x^2\)−36x+56=04x2−36x+56
⇒4(x2−9x+14)=0⇒4(x2−9x+14)
⇒4(x2−7x−2x+14)=0⇒4(x2−7x−2x+14)
⇒4x(x−2)−7(x−2)=0⇒4x(x−2)−7(x−2)
⇒4(x−7)(x−2)=0⇒4(x−7)(x−2)
⇒(x−7)(x−2)=0⇒(x−7)(x−2)
⇒[x−7=0x−2=0⇒[x−7=0x−2=0
⇒x=7;x=2⇒x=7;x=2.
a) \(x^3\left(x^2-7\right)^2-36x=x\left[\left(x^3-7x\right)^2-6^2\right]\)
\(=x\left(x^3-7x-6\right)\left(x^3-7x+6\right)\)
\(x\left[\left(x-3\right)\left(x+1\right)\left(x+2\right)\right].\left[\left(x+3\right)\left(x-2\right)\left(x-1\right)\right]\)
\(=\left(x-3\right)\left(x-2\right)\left(x-1\right).x.\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b) Không pt được.
c) Không pt được.
\(=x^2-36x+324-268\\ =\left(x-18\right)^2-268\\ =\left(x-18-2\sqrt{67}\right)\left(x-18+2\sqrt{67}\right)\)
\(4x^4+36x^2+81-36x^2=\left(4x^4+36x^2+81\right)-36x^2=\left(2x^2+9\right)^2-36x^2=\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)\)
\(4x^4+36x^2+81-36x^2\)
\(=\left(2x-9\right)^2-\left(6x\right)^2=\left(2x^2+9+6x\right)\left(2x^2-6x+9\right)=\left(2x^2+6x+9\right)\left(2x^2-6x+9\right)\)