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Xét số hạng tổng quát \(\frac{n+1}{n}=1+\frac{1}{n}\) . Vì \(0
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
\(\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]\right\}:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\)\(=\left\{\left[\left(2.2\right)^2:2,4\right]\left[5,25:\left(7\right)^2\right]\right\}:\left\{\left[\dfrac{15}{7}:\dfrac{\left(5\right)^2}{7}\right]\right\}:\left[4:\dfrac{\left(2.2\right)^2}{9}\right]\)
\(=\left\{\left[\left(4\right)^2:2,4\right]\left[5,25:49\right]\right\}:\left\{\left[\dfrac{15}{7}:\dfrac{25}{7}\right]\right\}:\left[4:\dfrac{\left(4\right)^2}{9}\right]\)
\(=\left\{\left[16:2,4\right].\dfrac{3}{28}\right\}:\left\{\dfrac{3}{5}\right\}:\left[4:\dfrac{8}{9}\right]\)
\(=\left\{\dfrac{20}{3}.\dfrac{3}{28}\right\}:\dfrac{3}{5}:\dfrac{9}{2}\)
\(=\dfrac{5}{7}:\dfrac{3}{5}:\dfrac{9}{2}\)
\(=\dfrac{5}{7}.\dfrac{5}{3}:\dfrac{9}{2}\)
\(=\dfrac{25}{21}:\dfrac{9}{2}\)
\(=\dfrac{25}{21}.\dfrac{2}{9}\)
\(=\dfrac{25.2}{21.9}\)
\(=\dfrac{50}{189}.\)
Mình làm chi tiết rồi nha bạn :))
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0