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a) Ta có: \(2x^2\left(3x^2-7x-5\right)\)
\(=2x^2\cdot3x^2-2x^2\cdot7x-2x^2\cdot5\)
\(=6x^4-14x^3-10x^2\)
c) Ta có: \(\left(16x^4-20x^2y^3-4x^5y\right):\left(-4x^2\right)\)
\(=16x^4:\left(-4x^2\right)+20x^2y^3:4x^2+4x^5y:4x^2\)
\(=-4x^3+5y^3+x^3y\)
\(=\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}=7x+y\)
\(\left[{}\begin{matrix}x=\dfrac{7-\sqrt{17}}{8}\\x=\dfrac{7+\sqrt{17}}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow x-2=0\) hay \(4x+1=0\)
\(\Leftrightarrow x=2\) \(\Leftrightarrow\) \(4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy \(S=\left\{2,-\dfrac{1}{4}\right\}\)
a, 3x2 - 8x2 - 2x+3=0
2x(3-8) - 2x+3=0
2x5 - 2x+3=0
2x5 - 2x=0-3=
2x5 - 2x=-3
2x(5-x)=-3
5-x=-3/2
5-x=1,5
x=5-1,5
x=3,5
\(4x^2-7x-2\\ =4x^2-8x+x-2\\ =4x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(4x+1\right)\)
\(=\left(4x^2-7x-50\right)^2-x^2\left(16x^2+56x+49\right)\)
\(=\left(4x^2-7x-50\right)^2-x^2\left(4x+7\right)^2\)
\(=\left(4x^2-7x-50-4x^2-7x\right)\left(4x^2-7x-50+4x^2+7x\right)\)
\(=\left(-14x-50\right)\left(8x^2-50\right)\)
\(=-4\left(7x+25\right)\left(2x-5\right)\left(2x+5\right)\)