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\(\left(\dfrac{1}{38}-1\right).\left(\dfrac{1}{37}-1\right).\left(\dfrac{1}{36}-1\right)....\left(\dfrac{1}{2}-1\right)\)
\(=\left(\dfrac{1}{38}-\dfrac{38}{38}\right).\left(\dfrac{1}{37}-\dfrac{37}{37}\right).\left(\dfrac{1}{36}-\dfrac{36}{36}\right)....\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\)
\(=\left(\dfrac{-37}{38}\right).\left(\dfrac{-36}{37}\right).\left(\dfrac{-35}{36}\right)...\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(\dfrac{-1}{1}\right).\left(\dfrac{-1}{1}\right).....\left(-\dfrac{1}{1}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(-1\right).\left(-1\right).....\left(-1\right)\)
\(=\dfrac{\left(-1\right).\left(-1\right).....\left(-1\right)}{38}\)
\(=\dfrac{\left(-1\right)^{38}}{38}\)
\(=\dfrac{1}{38}\)
\(\frac{1}{4}=\frac{1}{2.2}< \frac{1}{1.2}=\frac{1}{2}-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{16}< \frac{1}{2.4}=\frac{1}{4}-\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{36}< \frac{1}{4.6}=\frac{1}{8}-\frac{1}{12}\)
\(\Leftrightarrow\frac{1}{64}< \frac{1}{6.8}=\frac{1}{12}-\frac{1}{16}\)
\(\Leftrightarrow\frac{1}{100}< \frac{1}{8.10}=\frac{1}{16}-\frac{1}{20}\)
\(\Leftrightarrow\frac{1}{144}< \frac{1}{10.12}=\frac{1}{20}-\frac{1}{24}\)
\(\Leftrightarrow\frac{1}{196}< \frac{1}{12.14}=\frac{1}{24}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{16}+.....+\frac{1}{196}< \frac{1}{2}-\frac{1}{28}< \frac{1}{2}ĐPCM\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=-1.\)
b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=6.\frac{5}{4}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}.\)
c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)
\(=\frac{6}{7}.\)
d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)
\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)
\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)
\(=\frac{107}{100}+\frac{1}{5}\)
\(=\frac{127}{100}.\)
Chúc bạn học tốt!
a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{-59}{45}\)
b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)
\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)
\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)
\(\Rightarrow\frac{31}{4}\)
c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)
\(\Rightarrow\frac{6}{7}\)
d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)
\(\Rightarrow\frac{93}{100}\)
\(\text{}\text{}\)\(=\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}:\frac{43}{37}-\frac{6}{29}:\frac{41}{21}\\ =\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}.\frac{37}{43}-\frac{6}{29}.\frac{21}{41}\)
\(=\frac{-37}{38}\cdot\frac{-36}{37}\cdot...\cdot\frac{-1}{2}=\frac{\left(-37\right)\left(-36\right)\cdot...\cdot\left(-1\right)}{38\cdot37\cdot...\cdot2}=\frac{1}{38}\)