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a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)
\(=3x^2-4x-26-4x^2+16\)
\(=-x^2-4x-10\)
\(=\dfrac{3x\left(-x^2\right)}{3x}+\dfrac{2}{3x}-\dfrac{3x}{3x}=\dfrac{-3x^3+2-3x}{3x}\)
\(=\dfrac{-x^2+2-3x}{1}=-\left(x^2-2+3x\right)\)
vậy bt A luôn......
a: \(x^2-5x+10\)
\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{15}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\forall x\)
b: \(2x^2+8x+15\)
\(=2\left(x^2+4x+\dfrac{15}{2}\right)\)
\(=2\left(x^2+4x+4+\dfrac{7}{2}\right)\)
\(=2\left(x+2\right)^2+7>0\forall x\)
\(-9x^2+12x-15=\left(-11\right)-\left(9x^2-12x+4\right)=\left(-11\right)-\left(3x-2\right)^2\le-11< 0\)
\(-5-\left(x-1\right).\left(x+2\right)=-5-\left(x^2+x-2\right)=-\left(x^2+x+3\right)=-\left(\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\right)\le-\frac{11}{4}< 0\)
a: \(A=\dfrac{3\left(1-2x\right)}{2x\left(x^2+1\right)-\left(x^2+1\right)}\)
\(=\dfrac{-3\left(2x-1\right)}{\left(x^2+1\right)\left(2x-1\right)}=\dfrac{-3}{x^2+1}\)
b: Khi x=3 thì \(A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)
c: x^2+1>=0
=>3/x^2+1>=0
=>-3/x^2+1<=0
=>A<=0(ĐPCM)