\(\dfrac{9x-3}{4x^2+x}.\dfrac{16x^2+8x+1}{1-27x^3}=\dfrac{?}{x(1+3x+9x^2)}\)
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c) \(\dfrac{1}{\left(1-x\right)\left(x-2\right)}+\dfrac{1}{\left(2-x\right)\left(x-3\right)}+...+\dfrac{1}{\left(99-x\right)\left(x-100\right)}\)
\(=-\dfrac{1}{\left(x-1\right)\left(x-2\right)}-\dfrac{1}{\left(x-2\right)\left(x-3\right)}-...-\dfrac{1}{\left(x-99\right)\left(x-100\right)}\)
\(=-\left[\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-99\right)-\left(x-100\right)}{\left(x-99\right)\left(x-100\right)}\right]\)
\(=-\left(\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-100}-\dfrac{1}{x-99}\right)\)
\(=-\left(\dfrac{1}{x-100}-\dfrac{1}{x-1}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x-100}\)
Bài 5:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: \(x=-y\):
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{z}=\dfrac{1}{0+z}=\dfrac{1}{x+y+z}\)
Ta có đpcm.
Các trường hợp \(y=-z\) và \(z=-x\) tương tự.
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a) \(-5x^2+3x=-\left(5x^2-3x\right)=-\left[\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\dfrac{3}{2\sqrt{5}}+\left(\dfrac{3}{2\sqrt{5}}\right)^2-\dfrac{9}{20}\right]\)
\(=-\left(\sqrt{5}x-\dfrac{3}{2\sqrt{5}}\right)^2+\dfrac{9}{20}\le\dfrac{9}{20}\)
Dấu \(=\)xảy ra khi \(\sqrt{5}x-\dfrac{3}{2\sqrt{5}}=0\Leftrightarrow x=\dfrac{3}{10}\).
b) \(4x^2-3x=\left(2x\right)^2-2.\left(2x\right).\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\dfrac{9}{16}=\left(2x-\dfrac{3}{4}\right)^2-\dfrac{9}{16}\ge-\dfrac{9}{16}\)
Dấu \(=\) xảy ra khi \(2x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{8}\).
c) Bạn làm tương tự ý b).
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