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Bài 2:
a: \(=\dfrac{4x^2+3-19}{x-2}=\dfrac{4x^2-16}{x-2}=\dfrac{4\left(x-2\right)\left(x+2\right)}{x-2}=4x+8\)
b: \(=\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2}{x+2y}-\dfrac{1}{x-2y}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x-4y-x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{x-6y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
\(A=\left(x-y\right)\left(x^2-xy\right)-x\left(x^2+2y^2\right)\)
\(=x^3-x^2y-x^2y+xy^2-x^3-2xy^2\)
\(=-2x^2y-xy^2\)
\(=-2\cdot2^2\cdot\left(-3\right)-2\cdot\left(-3\right)^2\)
\(=8\cdot3-2\cdot9\)
=6
a: \(=15x^5y^3-6x^4y^2-6x^3y^3\)
c: \(=2x^4-2x^2-3x^3+3x+x^2-1\)
\(=2x^4-3x^3-x^2+3x-1\)
Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
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