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\(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(x^4-16y^4\)
\(=\left(x^2\right)^2-\left(4y^2\right)^2\)
\(=\left(x^2-4y^2\right)\left(x^2+4y^2\right)\)
\(=\left(x-2y\right)\left(x+2y\right)\left(x^2+4y^2\right)\)
Chúc bạn học tốt.
\(54x^3+16y^3=2\left(27x^3+8y^3\right)\)
\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2\left(3x+2y\right)\left[\left(3x\right)^2-3x.2y+\left(2y\right)^2\right]\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(54x^3+16y^3=2\left(27x^3+8y^3\right)=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(c,54x^3+16y^3=2.\left(27x^3+8y^3\right)=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(d,x^4-16y^4=\left(x^2\right)^2-\left(4y^2\right)^2=\left(x^2-4y^2\right)\)
a.\(x^2-64x=x\left(x-64\right)\)
b.\(24x^3-8=8\left(3x^3-1\right)\)
c.\(x^2-16y^2-3x+12y=\left(x^2-16y^2\right)-3\left(x-4y\right)\)\(=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)
k mình nha bn ^.^ thanks
16y2 - 4x2 - 12x - 9 = 16y2 - (4x2 + 12x + 9) = 16y2 - (2x + 3)2 = (4y - 2x - 3)(4y + 2x + 3)
\(16x^2+y^2+4y-16y-8xy\)
\(=\left(16x^2-8xy+y^2\right)+4y-16y\)
\(=\left(4x+y\right)^2-12y\)
\(=\left(4x+y-\sqrt{12y}\right)\left(4x+y-\sqrt{12y}\right)\)
P/S : Sai thì thôi nha!
\(A=x^4-6x^3+27x^2-54x+32\)
\(=x^4-5x^3+22x^2-32x-x^3+5x^2-22x+32\)
\(=x\left(x^3-5x^2+22x-32\right)-\left(x^3-5x^2+22x-32\right)\)
\(=\left(x-1\right)\left(x^3-5x^2+22x-32\right)\)
\(=\left(x-1\right)\left(x^3-3x^2+16x-2x^2+6x-32\right)\)
\(=\left(x-1\right)\left[x\left(x^2-3x+16\right)-2\left(x^2-3x+16\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-3x+16\right)\)
Vì \(x\in Z\)=> x-1;x-2 là 2 số nguyên liên tiếp => \(\left(x-1\right)\left(x-2\right)⋮2\)
\(\Rightarrow A=\left(x-1\right)\left(x-2\right)\left(x^2-3x+16\right)⋮2\) hay A là số chẵn (đpcm)
\(A=x^4-6x^3+27x^2-54x+32\)
\(=x^4-x^3-5x^3+5x^2+22x^2-22x-32x+32\)
\(=\left(x-1\right)\left(x^3-5x^2+22x-32\right)\)
\(=\left(x-1\right)\left[x^2\left(x-2\right)-3x\left(x-2\right)+16\left(x-2\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-3x+16\right)\)
Vì \(\left(x-1\right)\left(x-2\right)⋮2\) nên A là số chẵn với mọi x thuộc Z
16y2-x2-2x-1
=16y2-(x2+2x+1)
=(4y)2-(x+1)2
=[4y-(x+1)][4y+(x+1)]
=(4y-x-1)(4y+x+1)
\(16y^2-x^2-2x-1=\left(4y\right)^2-\left(x^2+2x+1\right)=\left(4y\right)^2-\left(x+1\right)^2=\left(4y-x-1\right)\left(4y+x+1\right)\)
Ta có: y^4 - 16y^2= (y^2)^2 - (4y)^2
=(y^2 - 4y).(y^2 + 4y)
\(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2.\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2.\left[\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\right]\)
Chúc bạn học tốt.
\(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)
\(=2\left(2y+3x\right)\left(4y^2-6xy+9x^2\right)\)