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a) \(\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{2}+\sqrt{3}\)
b) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)\(=2\sqrt{5}\)
d) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}=\sqrt{2}-\sqrt{3}-1-\sqrt{3}\)
\(=\sqrt{12}-\sqrt{2}-1\)
e) \(\sqrt{\left(\sqrt{3-1}^2\right)-\sqrt{3}}=\sqrt{\sqrt{2}^2-\sqrt{3}}=\sqrt{2-\sqrt{3}}\)
P/S: Ko chắc
Bài 1:
a: \(-\sqrt{\left(-3\right)^2}=-\left|-3\right|=-3\)
b: \(-\sqrt{\left(-2\right)^4}=-\left|\left(-2\right)^2\right|=-4\)
c: \(=-\sqrt{5^2}=-\left|5\right|=-5\)
d: \(=\sqrt{\left(-3\right)^6}=\sqrt{3^6}=\left|3^3\right|=27\)
e: \(-\sqrt{\left(-1\right)^8}=-\left|\left(-1\right)^4\right|=-1\)
a) \(\sqrt{25}+\sqrt{9}-\sqrt{16}\) = \(\sqrt{5^2}+\sqrt{3^2}-\sqrt{4^2}\) = 5 + 3 - 4 = 4
b) \(\sqrt{0,16}+\sqrt{0,01}+\sqrt{0,25}\) = 0,4 + 0,1 + 0,5 = 1
c) \(\left(\sqrt{3^2}\right)-\left(\sqrt{2^2}\right)+\left(\sqrt{5^2}\right)\)
= 3 - 2 + 5 = 6
d) \(\sqrt{4}-\left(-\sqrt{3}\right)^2+\sqrt{49}\) = 2 - 3 + 7 = 6
e) \(\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2\)
= \(\left(\sqrt{8}\right)^2-\left(\sqrt{27}\right)^2\) = 8 - 27 = -19
f) \(\left(-2\sqrt{2}\right)^2+\left(3\sqrt{3}\right)^2\) = 8 + 27 = 35
\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)
\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge2ac+2bd\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)
BĐT cuối đúng theo BĐT Bunhiacopski
Dấu "=" khi \(\frac{a}{c}=\frac{b}{d}\)
B1:
1. \(\sqrt{12.5}\cdot\sqrt{0.2}\cdot\sqrt{0.1}\) \(=\sqrt{12.5\cdot0.2\cdot0.1}\) \(=\sqrt{0.25}=0.5\)
2.\(\sqrt{48.4}\cdot\sqrt{5}\cdot\sqrt{0.5}\) = \(\sqrt{48.4\cdot5\cdot0.5}\) =\(\sqrt{121}=11\)
B2:
a, \(\left(\sqrt{7}+\sqrt{3}\right)^2=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3=7+2\cdot\sqrt{21}+3\)\(=10+2\sqrt{21}\)
b,\(\left(\sqrt{11}-\sqrt{5}\right)^2=11-2\sqrt{55}+5=16-2\sqrt{55}\)
c,\(\left(\sqrt{x}+\sqrt{y}\right) ^2=x+2\sqrt{xy}+y\)
d.\(\left(\sqrt{13}+\sqrt{7}\right)^2=13+2\sqrt{7}+7=20+2\sqrt{7}\)
e,\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)
f,\(\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\)
a: \(=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)
b: =8
c: \(=2-\sqrt{3}\)
d: \(=3-2\sqrt{2}\)
e: \(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)