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2)
\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)
\(=\sqrt{12,1.36.10}\)
= \(\sqrt{121.36}\)
\(=\sqrt{4356}\)
\(=66\)
3)
\(\sqrt{5a}.\sqrt{45a}-3a\)
\(=\sqrt{5.45a^2}-3a\)
\(=\sqrt{225a^2}-3a\)
\(=\sqrt{\left(15a\right)^2}-3a\)
\(=-15a-3a\) ( vì \(a\le0\))
\(=-18a\)
5)
\(\sqrt{0,36a^2}\)
\(=\sqrt{\left(0,6a\right)^2}\)
\(=-0,6a\) ( vì \(a< 0\) )
Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.
Chúc bạn học tốt!
1)
\(\sqrt{3a^3}.\sqrt{12}\)
\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)
\(=\sqrt{3.12}.\sqrt{a^3}\)
\(=6\sqrt{a^3}\)
4)
\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)
\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)
\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)
\(=54a^3-6\sqrt{a^2}\)
\(=54a^3-6a^2\) ( vì a<0)
6)
\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)
\(=\sqrt{\left(a^2\right)^2.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\right)^2}.\sqrt{\left(3-a\right)^2}\)
\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)
Mà\(\left|3-a\right|=-\left(-3-a\right)=-3+a=a-3\)
\(=a^2\left(a-3\right)\)
\(=a^3-3a^2\)
Còn lại bạn làm tương tự nha, trể quá rùi :)))))
a,\(\sqrt{4\left(a-5\right)^2}=\sqrt{4}.\sqrt{\left(a-5\right)^2}=2.\left|a-5\right|=2\left(a-5\right)\left(a\ge5\right)\)
b,\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3=-1}\)
c,Mạn phép sửa đề ,nếu ko thì kết quả ko đẹp
\(\sqrt{8+2\sqrt{15}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{5}=\sqrt{5}+\sqrt{3}-\sqrt{5}=\sqrt{3}\)
d,\(\sqrt{\left(3-2\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt{3}-3-3-2\sqrt{3}=-6\)
e,\(\sqrt{24\left(b-3\right)}^2=\sqrt{24^2}.\sqrt{\left(b-3\right)^2}=24.\left(3-b\right)\left(b< 3\right)\)
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
a) Ta có:
\(6\sqrt{5}=\sqrt{5\cdot36}=\sqrt{180}\)
\(5\sqrt{6}=\sqrt{6\cdot25}=\sqrt{200}\)
Mà \(\sqrt{180}< \sqrt{200}\)
Vậy: \(6\sqrt{5}< 5\sqrt{6}\)
x) Ta có: \(\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\)
Công hai vế của BĐT cho 3:
Suy ra: \(\sqrt{8}+3< 3+3=6\)
Vậy: \(\sqrt{8}+3< 6\)
b) Ta có:
\(\sqrt{2\sqrt{3}}=\sqrt[4]{12}\)
Tương tự: \(\sqrt{3\sqrt{2}}=\sqrt[4]{18}\)
Mà \(\sqrt[4]{18}>\sqrt[4]{12}\)
Vậy.....
d) Ta có:
\(2\sqrt{5}-5=\sqrt{5}+\sqrt{5}-5=\left(\sqrt{5}-2\right)+\left(\sqrt{5}-3\right)>\sqrt{5}-3\)
Vậy ......
e) Ta có:
\(\sqrt{2}-2=\frac{3\sqrt{2}-6}{3}\)
\(\sqrt{3}-3=\frac{2\sqrt{3}-6}{2}\)
Mà \(3\sqrt{2}>2\sqrt{3}\)
Vậy .....
f) ........... Đang thinking
a/ \(2\sqrt{6}=\sqrt{24};3\sqrt{3}=\sqrt{27}\)
Vì: 24 < 27 => \(\sqrt{24}< \sqrt{27}\)
hay \(2\sqrt{6}< 3\sqrt{3}\)
b/ \(\dfrac{2}{5}\sqrt{6}=\sqrt{\dfrac{24}{5}};\dfrac{7}{4}\sqrt{\dfrac{1}{3}}=\sqrt{\dfrac{49}{48}}\)
Có: \(\dfrac{24}{5}=\dfrac{1152}{240};\dfrac{49}{48}=\dfrac{245}{240}\)
Ta thấy: \(\dfrac{1152}{240}>\dfrac{245}{240}\Rightarrow\dfrac{24}{5}>\dfrac{49}{48}\)
\(\Rightarrow\sqrt{\dfrac{24}{5}}>\sqrt{\dfrac{49}{48}}\Rightarrow\dfrac{2}{5}\sqrt{6}>\dfrac{7}{4}\sqrt{\dfrac{1}{3}}\)
vậy........
a) \(3\sqrt[3]{2}=\sqrt[3]{3^3.2}=\sqrt[3]{54}>\sqrt[3]{53}\)
b) \(22=\sqrt[3]{22^3}=\sqrt[3]{10648}>\sqrt[3]{10638}=3\sqrt[3]{394}\)