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a; \(\sqrt{27a}\cdot\sqrt{3a}=\sqrt{81a^2}=9a\)

b: \(\dfrac{\sqrt{8a^4b^6}}{\sqrt{64a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=\dfrac{-\sqrt{2}}{4a}\)(do a<0)

10 tháng 9 2017

a) \(2\sqrt{a^2}=2\left|a\right|=2a\) (vì \(a\ge0\))

b) \(\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) (vì \(a< 0\))

c) \(5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)

d) \(\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)=-\dfrac{1}{3}c^3\) (vì \(c< 0\Rightarrow c^3< 0\))

10 tháng 9 2017

\(a)2\sqrt{a^2}=2.\left|a\right|=2a\) ( vì \(a\ge0\) )

\(b)\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) ( vì \(a< 0\) )

\(c)5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)

\(d)\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)\) ( vì \(c< 0\Rightarrow c^3< 0\) )

Chúc bn học tốt!

31 tháng 12 2023

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

20 tháng 10 2023

a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)

\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)

\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)

\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)

\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)

b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)

\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)

\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)

Bài 1: 

a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)

=>2 căn x=6

=>căn x=3

=>x=9

b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)

=>x=1

23 tháng 8 2021

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)

a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)

=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)

=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)

=>\(-2x=\dfrac{1}{4}\)

=>\(2x=-\dfrac{1}{4}\)

=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)

b: ĐKXĐ: x>=0

\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)

=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

4 tháng 12 2023

bài nào cũng thấy Phước Thịnh :)