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Đàooooo
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Nguyễn Lê Phước Thịnh
29 tháng 8 2021 lúc 14:58

Bài 1:

a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=8\sqrt{7}\)

Nguyễn Lê Phước Thịnh
30 tháng 8 2021 lúc 0:35

Bài 3: 

a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)

b: \(\sqrt{37^2-35^2}=12\)

c: \(\sqrt{65^2-63^2}=16\)

d: \(\sqrt{117^2-108^2}=45\)

Nguyễn Bùi Diễm Phúc
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Phạm Thị Thùy Linh
26 tháng 6 2019 lúc 9:04

\(a,\sqrt{\frac{5.\left(38^2-17^2\right)}{8.\left(47^2-19^2\right)}}\)

\(=\sqrt{\frac{5.\left(38-17\right)\left(38+17\right)}{8.\left(47-19\right)\left(47+19\right)}}\)

\(=\sqrt{\frac{5.21.55}{8.28.66}}\)

\(=\sqrt{\frac{5775}{14784}}=\frac{5\sqrt{231}}{2\sqrt{4370}}\)

Nguyễn Bùi Diễm Phúc
26 tháng 6 2019 lúc 9:08

.bn tính lại \(\sqrt{14784}\)đi sao lạ vậy

Nguyễn Thị Hà Uyên
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Huong San
11 tháng 9 2018 lúc 17:30

\(A=\dfrac{5.\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}\\ =\dfrac{5\left(38-17\right)\left(38+17\right)}{8\left(47-19\right)\left(47+19\right)}\\ =\dfrac{5.21.55}{8.28.66}\\ =\dfrac{5.1155}{8.1848}\\ =\dfrac{5.5}{8.8}\\ =\dfrac{25}{64}\)

\(B=\sqrt{\dfrac{0,2\times1,21\times0,3}{7,5\times3,2\times0,64}}\\ =\sqrt{0,0625\times1,890625\times0,04}\\ =\sqrt{\dfrac{121}{25600}}\\ =\dfrac{11}{160}\)

Khánh Linh
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Nguyễn Lê Phước Thịnh
13 tháng 5 2022 lúc 22:32

 

1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)

\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)

2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)

\(=-1+2=1\)

Nguyễn Bùi Diễm Phúc
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Trần Minh Hoàng
26 tháng 6 2019 lúc 9:32

Ta có:

\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\sqrt{3}+4\\x_2=-\sqrt{3}+2\end{matrix}\right.\)

Vậy...

Nguyen Diana
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Nguyễn Hoàng Minh
31 tháng 10 2021 lúc 16:49

\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)

Phạm Ninh Đan
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Nguyễn Lê Phước Thịnh
15 tháng 2 2021 lúc 12:12

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

\(=4-1-\dfrac{2}{5}\)

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

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

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

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

Nguyễn Châu Mỹ Linh
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Mạnh Ngoáy
11 tháng 11 2018 lúc 10:07

e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)

=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)

=\(-\dfrac{516}{7}\)

Mạnh Ngoáy
11 tháng 11 2018 lúc 9:17

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)

=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)

=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)

=\(\dfrac{119}{240}\)

Mạnh Ngoáy
11 tháng 11 2018 lúc 9:26

b)\(\dfrac{3}{2}-\dfrac{5}{6}:\left(\dfrac{1}{2}\right)^2+\sqrt{4}\)

=\(\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+2\)

=\(\dfrac{3}{2}-\dfrac{10}{3}+2\)

=\(-\dfrac{11}{6}\) +2

=\(\dfrac{1}{6}\)

Qúy Công Tử
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Mai Anh Phạm
3 tháng 1 2019 lúc 13:12

1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)

2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)

3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2} \)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)

4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)

5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)

7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)

8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2


Hương Trương
4 tháng 1 2019 lúc 12:47
https://i.imgur.com/pmexRQv.jpg
Hương Trương
4 tháng 1 2019 lúc 12:48
https://i.imgur.com/Fbx1rk4.jpg