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Ta thấy các số trong căn bậc hai đều lớn hơn 0, áp dụng \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)
a) \(\sqrt{7}\cdot\sqrt{63}=\sqrt{7\cdot63}=21\)
b) \(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30\cdot48}=60\)
c) \(\sqrt{0,4}\cdot\sqrt{6,4}=\sqrt{0,4\cdot6,4}=1,6\)
d) \(\sqrt{2,7}\cdot\sqrt{5}\cdot\sqrt{1,5}=\sqrt{2,7\cdot5\cdot1,5}=4,5\)
a. \(\sqrt{7}.\sqrt{63}=\sqrt{7.63}=\sqrt{441}=21\)
b.\(\sqrt{2,5}.\sqrt{30}.\sqrt{48}=\sqrt{2,5.30.48}=\sqrt{3600}=60\)
c.\(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)
d.\(\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{2,7.5.1,5}=\sqrt{20,25}=4,5\)
a) \(\sqrt{10}.\sqrt{40}\)
=\(\sqrt{10.40}\)
=\(\sqrt{400}\)
=20
b) \(\sqrt{5.}\sqrt{45}\)
=\(\sqrt{5.45}\)
=\(\sqrt{225}\)
=\(\sqrt{15}\)
c) \(\sqrt{52.}\sqrt{13}\)
=\(\sqrt{52.13}\)
=\(\sqrt{676}\)
=26
d)\(\sqrt{2.}\sqrt{162}\)
=\(\sqrt{2.162}\)
=\(\sqrt{324}\)
=18
a) \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\frac{4}{10}.\frac{64}{10}}=\sqrt{\frac{\left(2.8\right)^2}{10^2}}=\frac{16}{10}=\frac{8}{5}\)
b) \(\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{\frac{27}{10}.5.\frac{15}{10}}=\sqrt{\frac{3^3.5^2.3}{10^2}}=\sqrt{\frac{\left(3^2.5\right)^2}{10^2}}=\frac{45}{10}=\frac{9}{2}\)
câu này dễ mà
chỉ cần nhân vào là xong
kiến thức đầu lớp 9 khá dễ đấy
tự mình làm đi nha bạn
a)\(\sqrt{10}\cdot\sqrt{40}=\sqrt{10\cdot40}=\sqrt{400}=20\)
b) \(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{2\cdot2\cdot81}=\sqrt{4}\cdot\sqrt{81}=2\cdot9=18\)
a/ Đề sai
b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)
\(=-11\sqrt{5}+3\sqrt{2}\)
c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)
\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)
d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
Áp dụng quy tắc chia hai căn bậc hai, hãy tính :
a) 2300−−−−√23−−√ = \(\sqrt{\dfrac{2300}{23}}\) = \(\sqrt{100}\) = 10
b) 12,5−−−−√0,5−−−√ = \(\sqrt{\dfrac{12,5}{0,5}}\) = \(\sqrt{25}\) = 5
c) 192−−−√12−−√ = \(\sqrt{\dfrac{192}{12}}\) = \(\sqrt{16}\) = 4
d) 6–√150−−−√ = \(\sqrt{\dfrac{6}{150}}\) = \(\sqrt{\dfrac{1}{25}}\) = \(\dfrac{1}{5}\)
\(\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{\left(2\sqrt{2}-1\right)^2}}{2}=\frac{2\sqrt{2}-1}{2}\)
\(\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
\(\sqrt{\frac{289+4\sqrt{72}}{16}}=\frac{\sqrt{\left(12\sqrt{2}+1\right)^2}}{4}=\frac{12\sqrt{2}+1}{4}\)
\(\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1
a
\(\sqrt{3}\cdot\sqrt{75}=\sqrt{3\cdot75}=\sqrt{225}=15\)
b
\(\sqrt{72}\cdot\sqrt{18}=6\sqrt{2}\cdot3\sqrt{2}=18\cdot2=36\)
c
\(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30}\cdot\sqrt{48}=\sqrt{75}\cdot\sqrt{48}=5\sqrt{3}\cdot4\sqrt{3}=20\cdot3=60\)
d
\(\sqrt{\frac{5}{49}}\cdot\sqrt{\frac{16}{125}}=\sqrt{\frac{5}{49}\cdot\frac{16}{125}}=\sqrt{\frac{16}{49\cdot25}}=\frac{4}{7\cdot5}=\frac{4}{35}\)