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a) \(\left(6x^3+3x^2+4x+2\right):\left(3x^2+2\right)\)
\(=\left[3x^2\left(2x+1\right)+2\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=\left[\left(3x^2+2\right)\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=2x+1\)
b) \(\left(2x^3-22x^2-5x^2+60x+55x-150\right):\left(x-5\right)\)
\(=\left[\left(2x^3-22x^2+60x\right)-\left(5x^2-55x+150\right)\right]:\left(x-5\right)\)
\(=\left[2x\left(x^2-11x+30\right)-5\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-5x-6x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x-5\right)\left(x-6\right)\right]:\left(x-5\right)\)
\(=\left(2x-5\right)\left(x-6\right)\)
\(=2x^2-17x+30\)
d) \(\left(x^5+4x^3+3x^2-5x+15\right):\left(x^3-x+3\right)\)
\(=\left(x^5+5x^3+3x^2-x^3-5x+15\right):\left(x^3-x+3\right)\)
\(=\left[\left(x^5-x^3+3x^2\right)+\left(5x^3-5x^2+15\right)\right]:\left(x^3-x+3\right)\)
\(=\left[x^2\left(x^3-x+3\right)+5\left(x^3-x^2+3\right)\right]:\left(x^3-x+3\right)\)
\(=\left[\left(x^2+5\right)\left(x^3-x+3\right)\right]:\left(x^3-x+3\right)\)
\(=x^2+5\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
a) (x3 - 11x +5 - 3x2) : (x - 5)
= (x3 - 3x2 - 11x + 5) : (x - 5)
= (x3 - 5x2 + 2x2 - 10x - x + 5) : (x - 5)
= [(x3 - 5x2) + (2x2 - 10x) - (x - 5)] : (x - 5)
= [x2(x - 5) + 2x(x - 5) - (x - 5)] : (x - 5)
= (x2 + 2x - 1)(x - 5) : (x - 5)
= x2 + 2x - 1
b) (4x4 - 5x2 - 3 + 3x3 + 9x): (x2 - 3)
con b bn xem có chép nhầm đề bài ko nhé