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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) \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\frac{4}{10}.\frac{64}{10}}=\sqrt{\frac{\left(2.8\right)^2}{10^2}}=\frac{16}{10}=\frac{8}{5}\)
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) \(\sqrt{10}.\sqrt{40}\)
=\(\sqrt{10.40}\)
=\(\sqrt{400}\)
=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, \(\sqrt{0.09\cdot64=\sqrt{0.09}\cdot\sqrt{64}=0.3\cdot8=2.4}\)
b, \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{16\cdot49}=\sqrt{16}\cdot\sqrt{49}=4\cdot7=28\)
c, \(\sqrt{121\cdot360}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)
d, \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=18\)
a)\(\sqrt{0,09}.\sqrt{64}\)=0,3.8=2,4
b)\(\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)=4.7=28
c)\(\sqrt{121.36}\)=\(\sqrt{121}.\sqrt{36}\)=11.6=66
d)\(\sqrt{2^2}.\sqrt{3^4}\)=2.9=18
a, \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)do \(0,1>0\)
b, \(\sqrt{\left(-0,3\right)^2}=\sqrt{\left(0,3\right)^2}=\left|0,3\right|=0,3\)do \(0,3>0\)
c, \(-\sqrt{\left(-1,3\right)^2}=-\sqrt{\left(1,3\right)^2}=-\left|1,3\right|=-1,3\)do \(1,3>0\)
d, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\sqrt{\left(0,4\right)^2}=-0,4.\left|0,4\right|=-0,4.0,4=-0,14\)
do \(0,4>0\)
\(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)
\(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)
\(-\sqrt{\left(-1,3\right)^2}=-\left|-1,3\right|=-1,3\)
\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,16\)
a)\(\sqrt{10}\cdot\sqrt{40}=\sqrt{10\cdot40}=\sqrt{400}=20\)
b) \(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{2\cdot2\cdot81}=\sqrt{4}\cdot\sqrt{81}=2\cdot9=18\)
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
a) = \(\sqrt{10.40}=\sqrt{400}=\sqrt{20^2}=20\)
b) \(=\sqrt{5.45}=\sqrt{5^2.3^2}=\sqrt{15^2}=15\)
Học tốt nhé :)
a)\(\sqrt{7.63}\)=21
b)\(\sqrt{2,5.30.48}\)=60
c)\(\sqrt{0,4.6,4}\)=1,6
d)\(\sqrt{2,7.5.1,5}\)=4,5