Thự hiện phép tính:
\(\left(\left(2\sqrt{2}\right)^2:2,4\right).\left(5,25:\left(\sqrt{7}\right)^2\right):\left(\left(2\frac{1}{7}:\frac{\left(\sqrt{5}\right)^2}{7}\right):2^2.\frac{\left(2.\sqrt{2}\right)}{\sqrt{81}}^2\right)\)
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2 tháng 2 2022
\(=\left(8+2.4\right)\left(5.25:7\right):\left\{\left[\dfrac{15}{7}+\dfrac{5}{7}\right]:\left[4:\dfrac{8}{9}\right]\right\}\)
\(=10.4\cdot\dfrac{3}{4}:\left\{\dfrac{20}{7}:\dfrac{9}{2}\right\}\)
\(=7.8:\dfrac{40}{63}=12.285\)
H
24 tháng 12 2023
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
24 tháng 12 2023
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 \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
O
9 tháng 1 2018
\(\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]\right\}:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\)\(=\left\{\left[\left(2.2\right)^2:2,4\right]\left[5,25:\left(7\right)^2\right]\right\}:\left\{\left[\dfrac{15}{7}:\dfrac{\left(5\right)^2}{7}\right]\right\}:\left[4:\dfrac{\left(2.2\right)^2}{9}\right]\)
\(=\left\{\left[\left(4\right)^2:2,4\right]\left[5,25:49\right]\right\}:\left\{\left[\dfrac{15}{7}:\dfrac{25}{7}\right]\right\}:\left[4:\dfrac{\left(4\right)^2}{9}\right]\)
\(=\left\{\left[16:2,4\right].\dfrac{3}{28}\right\}:\left\{\dfrac{3}{5}\right\}:\left[4:\dfrac{8}{9}\right]\)
\(=\left\{\dfrac{20}{3}.\dfrac{3}{28}\right\}:\dfrac{3}{5}:\dfrac{9}{2}\)
\(=\dfrac{5}{7}:\dfrac{3}{5}:\dfrac{9}{2}\)
\(=\dfrac{5}{7}.\dfrac{5}{3}:\dfrac{9}{2}\)
\(=\dfrac{25}{21}:\dfrac{9}{2}\)
\(=\dfrac{25}{21}.\dfrac{2}{9}\)
\(=\dfrac{25.2}{21.9}\)
\(=\dfrac{50}{189}.\)
Mình làm chi tiết rồi nha bạn :))
25 tháng 9 2021
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
25 tháng 9 2021
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
12 tháng 10 2020
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
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