Tính:
a) \(\sqrt{\left(0,1\right)^2}\)
b) \(\sqrt{\left(-0,3\right)^2}\)
c) \(-\sqrt{\left(-1,3\right)^2}\)
d) \(-0,4\sqrt{\left(-0,4\right)^2}\)
Tính:
a. \(\sqrt{\left(0,1\right)^2};\) b. \(\sqrt{\left(-0,3\right)^2};\) c. \(-\sqrt{\left(-1,3\right)^2};\) d. \(-0,4\sqrt{\left(-0,4\right)^2}.\)
a, 0,1
b,0,3
c,-1,3
d,-0,16
Tính:
\(a,\sqrt{0,1^2}\)
\(b,\sqrt{\left(-0,4\right)^2}\)
\(c,-\sqrt{\left(-1,7\right)^2}\)
\(d,-0,5\sqrt{\left(-0,5\right)^4}\)
\(e,\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(g,\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(a,\sqrt{0,1^2}=0,1\)
\(b,\sqrt{\left(-0,4\right)^2}=|-0,4|=0,4\)
\(c,-\sqrt{\left(-1,7\right)^2}=-|-1,7|=-1,7\)
\(d,-0,5\sqrt{\left(-0,5\right)^4}=\frac{-1}{2}\sqrt{[\left(\frac{-1}{2}\right)^2]^2}=-\frac{1}{2}.\left(\frac{1}{2}\right)^2=\frac{-1}{2}.\frac{1}{4}=\frac{-1}{8}\)
\(e,\sqrt{\left(1-\sqrt{2}\right)^2}=|1-\sqrt{2}|=\sqrt{2}-1\)
\(g,\sqrt{\left(\sqrt{3}-1\right)^2}=|\sqrt{3}-1|=\sqrt{3}-1\)
i,\(\sqrt{12,1.360}\)
k,\(\sqrt{0,4}.\sqrt{6,4}\)
l,-0,4\(\sqrt{\left(-0,4\right)^2}\)
m,\(\sqrt{2^4.\left(-7\right)^2}\)
\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)
\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)
\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)
\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)
i, \(\sqrt{12,1\cdot360}=\sqrt{4356}=\sqrt{66^2}=66\)
k, \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4\cdot6,4}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{2^6}{5^2}}=\dfrac{2^3}{5}=\dfrac{8}{5}\)
l, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,4\cdot0,4=-\dfrac{4}{25}\)
m, \(\sqrt{2^4\cdot\left(-7\right)^2}=2^2\cdot\left|-7\right|=4\cdot7=28\)
i) \(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}=\left(\sqrt{12,1}.\sqrt{10}\right)\sqrt{36}=\sqrt{121}.\sqrt{36}=11.6=66\)
k) \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)
l) \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\left|-0,4\right|=-0,4.0,4=0,16\)
m) \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{2^2.2^2.\left(-7\right)^2}=\left|2.2.7\right|=28\)
Tính:
a) \(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5\)
b) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2\)
c) \(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)
d) \(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
e) \(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
* Tính:
a.\(\dfrac{-4}{3}.\sqrt{\left(-0,4\right)^2}\)
b.\(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
c.\(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)
\(=-\dfrac{4}{3}\cdot0.4\)
\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)
b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{7}\)
\(=\dfrac{6}{7}\)
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
f)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
g)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0,4}\right)\)
giải chi tiết cụ thể giúp mk với ạ
Hãy viết các số sau theo thứ tự tăng dần :
a) \(\left(0,3\right)^{\pi};\left(0,3\right)^{0,5};\left(0,3\right)^{\dfrac{2}{3}};\left(0,3\right)^{3,1415}\)
b) \(\sqrt{2^{\pi}};\left(1,9\right)^{\pi};\left(\dfrac{1}{\sqrt{2}}\right)^{\pi};\pi^{\pi}\)
c) \(5^{-2};5^{-0,7};5^{\dfrac{1}{3}};\left(\dfrac{1}{5}\right)^{2,1}\)
d) \(\left(0,5\right)^{-\dfrac{2}{3}};\left(1,3\right)^{-\dfrac{2}{3}};\pi^{-\dfrac{2}{3}};\left(\sqrt{2}\right)^{-\dfrac{2}{3}}\)
Tính:
a) \(\left(4+\sqrt{3+2}\right)+\sqrt{2-4\sqrt{3}}\)
b) \(\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\)
b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)
=27-20căn 5
a: 2-4căn 3<0
nên biểu thức ko có giá trị
\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)
`b)(4-2\sqrt5)^2-(\sqrt5+2)^2`
`=(4-2\sqrt5-\sqrt5-2)(4-2\sqrt5+\sqrt5+2)`
`=(2-3\sqrt5)(6-\sqrt5)`
$---------$
Áp dụng HĐT :
`a^2-b^2=(a-b)(a+b)`