1. \(\sqrt{3}\) - \(2\sqrt{12}\) \(3\sqrt{27}\) - \(4\sqrt{48}\) \(5\sqrt{75}\) 2. \(\sqrt{175}\) - \(2\sqrt{112}\) - \(3\sqrt{63}\) - \(2\sqrt{28}\) \(\sqrt{7}\) 3. \(3\sqrt{75}\)

KL

Khang Lạc

10 tháng 6 2019

1. \(\sqrt{3}\) - \(2\sqrt{12}\) + \(3\sqrt{27}\) - \(4\sqrt{48}\) + \(5\sqrt{75}\) 2. \(\sqrt{175}\) - \(2\sqrt{112}\) - \(3\sqrt{63}\) - \(2\sqrt{28}\) + \(\sqrt{7}\) 3. \(3\sqrt{75}\) - \(\frac{12}{\sqrt{2}}\) - \(30\sqrt{1\frac{1}{3}}\) + \(\sqrt{75}\) 4. (\(2\sqrt{3}\) -\(\sqrt{12}\) +\(2\sqrt{27}\) ) : \(3\sqrt{3}\) 5. (\(4\sqrt{2}\) - \(\sqrt{72}\) - \(3\sqrt{50}\) ) \(\sqrt{2}\) -\(4\sqrt{25}\) +...

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#Toán lớp 9

2

HT

Hoàng Tử Hà

10 tháng 6 2019

bấm máy tính là ra hết bn ak

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KL

Khang Lạc

10 tháng 6 2019

giải giúp mình đi bạn

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AT

An Tuệ

28 tháng 6 2018 - olm

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#Toán lớp 9

0

HP

Hoàng Phú Lợi

17 tháng 12 2023

\(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

\(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

17 tháng 12 2023

a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)

=0

b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

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

\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)

\(=\sqrt{3}+2-\sqrt{3}\)

=2

c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)

\(=4-\sqrt{7}+\sqrt{7}\)

=4

d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)

\(=16\sqrt{5}\)

e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)

\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

\(=-16\sqrt{3}\)

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AT

An Tuệ

28 tháng 6 2018 - olm

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#Toán lớp 9

0

NV

Nhĩ Vương Gia

22 tháng 8 2019

thực hiện các phép tính

\(a.\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)

b.\(\left(2\sqrt{112}-5\sqrt{7}+2\sqrt{63}-2\sqrt{28}\right)\sqrt{7}\)

c.\(\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)\)

d.\(7\sqrt{24}-\sqrt{150}-5\sqrt{54}\)

#Toán lớp 9

1

KH

Kiêm Hùng

22 tháng 8 2019

\(a.\\ \left(\sqrt{4.3}-\sqrt{16.3}-\sqrt{36.3}-\sqrt{64.3}\right)\\ =\left(2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):2\sqrt{3}\\ =\frac{-16\sqrt{3}}{2\sqrt{3}}=-8\)

\(b.\\ =\left(2\sqrt{16.7}-5\sqrt{7}+2\sqrt{9.7}-2\sqrt{4.7}\right)\sqrt{7}\\ =\left(8\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7}\right)\sqrt{7}\\ =5\sqrt{7}.\sqrt{7}=5.7=35\)

\(c.\\ =\left(2\sqrt{9.3}-3\sqrt{16.3}+3\sqrt{25.3}-\sqrt{64.3}\right)\left(1-\sqrt{3}\right)\\ =\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)\\ =\sqrt{3}\left(1-\sqrt{3}\right)\\ =\sqrt{3}-3\)

\(d.\\ =7\sqrt{4.6}-\sqrt{25.6}-5\sqrt{9.6}\\ =14\sqrt{6}-5\sqrt{6}-15\sqrt{6}=-6\sqrt{6}\)

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HY

Hải Yến Lê

23 tháng 8 2020

Bài tập:Rút gọn 1.\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) 2. \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\) 3....

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#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

23 tháng 8 2020

1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)

\(=1+\sqrt{2}\)

2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)

\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)

\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)

\(=\sqrt{3}\left(6-4+3\right)\)

\(=5\sqrt{3}\)

3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)

Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)

\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)

\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)

\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)

\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)

\(=2\sqrt{6}-12\sqrt{3}\)

4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)

\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)

\(=\frac{6-2\sqrt{3}}{6}\)

\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)

\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)

5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)

\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)

6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)

\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)

\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)

\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)

\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)

\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)

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AT

An Tuệ

28 tháng 6 2018 - olm

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#Toán lớp 9

0

NN

nguyet nguyen

17 tháng 7 2018

a, \(6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}\)

b, \(\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}\)

c,\(5\sqrt{44}-2\sqrt{275}-3\sqrt{176}\)

d, \(2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}\)

#Toán lớp 9

2

ND

Nhã Doanh

17 tháng 7 2018

\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=\left(6-4+50-63\right)\sqrt{3}=-11\sqrt{3}\)

\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=\left(2+9-30-\dfrac{6}{5}\right)\sqrt{7}=-20,2\sqrt{7}\)\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=10\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-12\sqrt{11}\)

\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)

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PK

Phùng Khánh Linh

17 tháng 7 2018

\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=-11\sqrt{3}\)

\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=-\dfrac{101}{5}\sqrt{7}\)

\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=20\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-2\sqrt{11}\)

\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)

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TT

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