Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\sqrt{12,5}}{\sqrt{0,5}}=\frac{\sqrt{0,5}.\sqrt{25}}{\sqrt{0,5}}=\sqrt{25}=5\)
\(\frac{\sqrt{6}}{\sqrt{150}}=\frac{\sqrt{6}}{\sqrt{6}.\sqrt{25}}=\frac{1}{\sqrt{25}}=\frac{1}{5}\)
\(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8+3,2\right)\left(6,8-3,2\right)}=\sqrt{10.3,6}=\sqrt{36}=6\)
\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8+18,2\right)\left(21,8-18,2\right)}=\sqrt{40.3,6}=\sqrt{144}=12\)
\(\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=5\)
\(\frac{\sqrt{6}}{\sqrt{150}}=\sqrt{\frac{6}{150}}=\sqrt{\frac{1}{25}}=\frac{1}{5}\)
\(\sqrt{\left(6,8\right)^2-\left(3,2\right)^2}=\sqrt{46,24-10,24}=\sqrt{36}=6\)
\(\sqrt{\left(21,8\right)^2-\left(18,2\right)^2}=\sqrt{475,24-331,24}=\sqrt{144}=12\)
a) \(\sqrt{6,8^2-3.2^2}\)
\(=\sqrt{\left(6,8+3,2\right).\left(6,8-3,2\right)}\)
\(=\sqrt{3,6.10}=\sqrt{36}=6\)
b) \(\sqrt{21,8^2-18,2^2}\)
\(=\sqrt{\left(21,8-18,2\right).\left(21,8+18,2\right)}\)
\(=\sqrt{3,6.40}=\sqrt{4.36}=2.6=12\)
\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)
a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)
=\(\sqrt{3,6.10}=\sqrt{36}=6\)
b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)
=\(\sqrt{3,6.40}=\sqrt{144}=12\)
c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)
d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)
Câu 1:
a, \(\sqrt{50.98} = 5\sqrt{2} . 7\sqrt{2} = 70\)
b, \(\sqrt{2,5.12,1} = 30,25\)
c, \(\sqrt{17.51.27} = \sqrt{23409} = 153\)
d, \(\sqrt{32.128} = \sqrt{4096} = 64\)
e, \(\sqrt{3,2.7,2.49} = 7\sqrt{3,2.7,2} = 7\sqrt{23,04} =33,6\)
g, \(\sqrt{2,5.12,5.20} = \sqrt{625} = 25\)
1. 95002
2. 444240
3.262440000
4.1494093039
6.3315710
7.3673076923/100000000
9.2111336 cm2
10. x - 18,6 = 98,2 + 6,8
x - 18,6 = 105
x = 105 + 18,6
x = 123,6
Vậy x = 123,6
a: \(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\cdot10\sqrt{2}\right)\cdot8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}\)
\(=54\sqrt{2}\)
b: \(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)
c: \(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
d: \(=\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}=0\)
bài 2 rút gọn :
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
= \(\left|1-\sqrt{2}\right|+\left|\sqrt{2}-3\right|\)
=\(\sqrt{2}-1+3-\sqrt{2}\)
=2
b) \(\sqrt{4-2\sqrt{3}}+\sqrt{7}-\sqrt{48}\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{3}-1+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{7}-3\sqrt{3}+1\)
c)
Lời giải:
$\sqrt{6,8^2-3,2^2}=\sqrt{(6,8-3,2)(6,8+3,2)}=\sqrt{3,6.10}=\sqrt{36}=6$
Akai Haruma Em xin lỗi nhưng có thể giúp e bài này đc k ạ
Câu hỏi của Hàn Thất - Toán lớp 8 | Học trực tuyến