\(\sqrt{49}+\sqrt{\left(-5\right)^2}-5\sqrt{1,44}+3\sqrt{\f...">
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17 tháng 9 2019

\(\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2+\left(4\sqrt{0,5}\right)^2-\left(\frac{1}{5}\sqrt{125}\right)^2\)

\(=2^2.3-3^2.2+4^2.0,5-5\)

\(=12-18+8-5\)

\(=-3\)

17 tháng 9 2019

                                                           Bài giải

\(\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2+\left(4\sqrt{0,5}\right)^2-\left(\frac{1}{5}\sqrt{125}\right)^2\)

\(=2^2\cdot3-3^2\cdot2+4^2\cdot0,5-\frac{1}{25}\cdot125\)

\(=12-18+8-5\)

\(=-3\)

5 tháng 10 2018

4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)

\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)

Tìm z thì dễ rồi

24 tháng 7 2017

a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}\)

\(=5+5-3-3\)

\(=4\)

b) \(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)

\(=\left(4+4\right).\frac{1}{8}-\frac{1}{9}\)

\(=8.\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)

\(=-4:\dfrac{13}{12}\)

\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)

e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)

=20-12+5-6

=8+5-6

=13-6=7

f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)

\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)

\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)

29 tháng 10 2020

a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)

\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)

\(=\frac{4}{5}+11-2\)

\(=\frac{4}{5}+9\)

\(=\frac{49}{9}\)

b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)

\(=-8+4-5+64\)

= 55

c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)

\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)

\(=\frac{\sqrt{48}}{91-7}\)

\(=\frac{4\sqrt{3}}{84}\)

\(=\frac{\sqrt{3}}{41}\)

d) Xem lại đề nhé em!

e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)

\(=5-3.\frac{2}{3}\)

= 5 - 2

= 3

h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)

\(=-9.\frac{1}{3}-7+125:5\)

\(=-3-7+25\)

= 15

26 tháng 2 2020

Bài 1:

a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)

\(=1-1\)

\(=0.\)

b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)

\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(-1\right)+1:\frac{4}{5}\)

\(=\left(-1\right)+\frac{5}{4}\)

\(=\frac{1}{4}.\)

c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)

\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)

\(=-\frac{\sqrt{205}}{4}.\)

d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)

\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)

\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)

\(=\frac{1}{9}.\frac{20}{11}\)

\(=\frac{20}{99}.\)

Chúc bạn học tốt!

26 tháng 2 2020

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