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a: \(=\dfrac{-4}{5}\cdot\left|-\dfrac{5}{4}\right|=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)
b: \(=\left|\left(-2\right)^3\right|=8\)
c: \(=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d: \(=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
e: \(=\left|\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\right|=\dfrac{\sqrt{2}-1}{2}\)
f: \(=\left|\dfrac{1}{10}-\dfrac{\sqrt{10}}{10}\right|=\dfrac{\sqrt{10}-1}{10}\)
a) \(-0,8\sqrt{\left(-0,125\right)^2}=-0,8.\left|-0,125\right|=-0.8.0,125=-\dfrac{1}{10}\)
b) \(\sqrt{\left(-2\right)^6}=\sqrt{\left(\left(-2\right)^3\right)^2}=\left|\left(-2\right)^3\right|=8\)
c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
a, \(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|3\sqrt{2}-2\right|\)
\(=\sqrt{2}-1+3\sqrt{2}-2=4\sqrt{2}-3\)
b, \(2\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}+1\right)^2}\)
\(=2\left|\sqrt{3}-1\right|-\left|2\sqrt{3}+1\right|\)
\(=2\sqrt{3}-2-2\sqrt{3}-1=-3\)
a)
\(\left(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =\left(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =3-2\\ =1\)
b)
\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)
a: \(=\left(\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
=(căn 3-căn 2)(căn 3+căn 2)
=3-2=1
b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)
=(2-căn 11)(2+căn 11)
=4-11
=-7
\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)
\(a,\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\)
\(=\sqrt{5}-2-\left(1+\sqrt{5}\right)\)
\(=\sqrt{5}-2-1-\sqrt{5}\)
\(=-3\)
\(b,\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+6\cdot\dfrac{\sqrt{4}}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12\sqrt{3}}{3}\)
\(=\sqrt{3}+4\sqrt{3}\)
\(=5\sqrt{3}\)
#\(Toru\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\\ =\sqrt{5}-2-1-\sqrt{5}\\ =-2-1\\ =-3\)
\(\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+4\sqrt{3}\\ =\sqrt{3}+4\sqrt{3}\\ =5\sqrt{3}\)
a: \(=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)
b: =8
c: \(=2-\sqrt{3}\)
d: \(=3-2\sqrt{2}\)
e: \(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)