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Ta có :
\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+3x^2+3x+1+5x^2+14x+9}\) \(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+14x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+5x+9x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x\left(x+1\right)+9\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x-2\right)}{\left(x+1\right)^3+\left(x+1\right)\left(5x+9\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+7x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+2x+5x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}\)
\(=\dfrac{x-2}{x+5}\)
\(\Rightarrow a=-2;b=5\)
Vậy a + b = -2 + 5 = 3
Ta có 3x^2-x+1=3x^2+2x-3x-2+3=(3x-2)(x-1)+3
D có giá trị nguyên\(\) khi 3\(⋮\)(3x+2)\(\Leftrightarrow\)3x+2 là ước của 3\(\Leftrightarrow\)3x+2\(\in\){-3;-1;1;3} suy ra x\(\in\){-5/3;-1;-1/3;1/3}mà x nguyên nên ta tìm được x=-1
\(2x+3y+5z=\frac{x^2+y^2+z^2}{2}+19\)
\(x^2+y^2+z^2+38=4x+6y+10z\)
\(\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\left(x-2\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
\(x-2=y-3=z-5=0\)
\(x=2,y=3,z=5\)