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29 tháng 12 2021

a, =2.0,1-0,5                                  b,=\(\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

    =0,2-0,5                                        =  -10:(\(\dfrac{-5}{7}\))

    =-0,3                                             = 14

           c, \(=\dfrac{5^4.\left(4.5\right)^4}{\left(5^2\right)^4.4^5}=\dfrac{5^4.4^4.5^4}{5^8.4^5}=\dfrac{1}{4^{ }}\)

16 tháng 12 2018

a, \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)

\(=15\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)-25\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)\)

\(=-\dfrac{7}{5}.\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\)

\(=-\dfrac{7}{5}.\left(-10\right)\)

\(=\dfrac{7}{5}.10\)

\(=\dfrac{7}{1}.2\)

\(=14\)

b, \(\sqrt{0.16}-\sqrt{0.25}\)

\(=0.4-0.5\)

\(=-0.1\)

`#3107.101107`

a)

`2/5 \sqrt{25} - 1/2 \sqrt{4}`

`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`

`= 2/5*5 - 1/2*2`

`= 2 - 1`

`= 1`

b)

`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`

`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`

`= 0,5*0,3 + 5*0,9`

`= 0,15 + 4,5`

`= 4,65`

c)

`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`

`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`

`= 2/5*5/6 - 5/2*2/5`

`= 1/3 - 1`

`= -2/3`

d)

`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`

`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`

`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`

`= -2*6/4 + 5*9/5`

`= -3 + 9`

`= 6`

5 tháng 10 2023

Xem lại kết quả câu c nhé bạn!

11 tháng 12 2021

b: =3/4x(-1)=-3/4

22 tháng 7 2023

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

22 tháng 7 2023

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

 

25 tháng 10 2022

a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)

b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)

\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)

c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)

 

17 tháng 12 2020

\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)

17 tháng 12 2020

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26 tháng 12 2023

a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

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

\(=-\dfrac{3}{2}\)

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)

\(=-1+\dfrac{5}{2}\)

\(=\dfrac{3}{2}\)

 

26 tháng 12 2023

56:54=

 

\(B=\dfrac{\left(\dfrac{5}{70}-\dfrac{10\sqrt{2}}{70}+\dfrac{6\sqrt{2}}{70}\right)\cdot\dfrac{-4}{15}}{\left(\dfrac{5}{50}+\dfrac{6\sqrt{2}}{50}-\dfrac{10\sqrt{2}}{50}\right)\cdot\dfrac{5}{7}}=\dfrac{\dfrac{5-4\sqrt{2}}{70}\cdot\dfrac{-4}{15}}{\dfrac{5-4\sqrt{2}}{50}\cdot\dfrac{5}{7}}\)

\(=\dfrac{-4\left(5-4\sqrt{2}\right)}{70\cdot15}\cdot\dfrac{50\cdot7}{5\left(5-4\sqrt{2}\right)}=\dfrac{-4}{5}\cdot\dfrac{350}{70\cdot15}=\dfrac{-4}{5}\cdot\dfrac{1}{3}=\dfrac{-4}{15}\)