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\(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
=X4-3X3 +6X3-18X2+11X2-33X+6X-18
=(X-3)(X3+6X2+11X+6)
=(X-3)(X+3)(X+1)(X+2)
\(x^4+3x^3-7x^2-27x-18.\)
\(=\left(x^4-9x^2\right)+\left(3x^3-27x\right)+\left(2x^2-18\right)\)
\(=x^2\left(x-3\right)\left(x+3\right)+3x\left(x-3\right)\left(x+3\right)+2\left(x-3\right)\left(x+3\right).\)
\(=\left(x-3\right)\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x+1\right)\left(x+2\right).\)
12x3 + 4x2 - 27x - 9
= 4x2 ( 3x + 1 ) - 9 ( 3x + 1 )
= ( 3x +1 ) ( 4x2 -9 )
k nha !
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
Chúc bạn học tốt.
\(A=x^4-6x^3+27x^2-54x+32\)
\(=x^4-5x^3+22x^2-32x-x^3+5x^2-22x+32\)
\(=x\left(x^3-5x^2+22x-32\right)-\left(x^3-5x^2+22x-32\right)\)
\(=\left(x-1\right)\left(x^3-5x^2+22x-32\right)\)
\(=\left(x-1\right)\left(x^3-3x^2+16x-2x^2+6x-32\right)\)
\(=\left(x-1\right)\left[x\left(x^2-3x+16\right)-2\left(x^2-3x+16\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-3x+16\right)\)
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
\(3z^2+6zy+3y^2-27x^2\)
\(=3\left(z^2+2zy+y^2-9x^2\right)\)
\(=3\left(\left(z+y\right)^2-\left(3x\right)^2\right)\)
\(=3\left(z+y-3x\right)\left(z+y+3x\right)\)