Odpowiedź:
[tex]\huge\boxed{a)~~\dfrac{\mid \sqrt{18} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid = 6\sqrt{2} -7}[/tex]
[tex]\huge\boxed{b)~~\dfrac{\mid 2-\sqrt{3} \mid \cdot \mid \sqrt{3}-2 \mid }{ \mid 2\sqrt{3}-5 \mid} =\dfrac{11-6\sqrt{3} }{13} }[/tex]
[tex]\huge\boxed{c)~~\dfrac{ \mid (\sqrt{7} -2)\cdot (\sqrt{7} +2)\mid}{\mid (1-\sqrt{3} )\cdot (\sqrt{3} -1)\mid } -\dfrac{\mid 3\sqrt{3} -5\mid }{2} =5\frac{1}{2} }[/tex]
[tex]\huge\boxed{d)~~\dfrac{\mid (3-\sqrt{6} )\cdot (3-\sqrt{6} )\mid }{\mid \sqrt{6} -2\mid \cdot \mid 2-\sqrt{6}\mid } +\dfrac{1}{2} \mid 6\sqrt{6} -11\mid =2}[/tex]
Szczegółowe wyjaśnienie:
Wartość bezwzględna:[tex]\huge\boxed{\mid x\mid = \left \{ {{x~~dla~~x\geq 0} \atop {-x~~dla~~x < 0}} \right\} }[/tex]
Korzystamy ze wzorów skróconego mnożenia:
Obliczamy wartość wyrażeń:
[tex]\huge\boxed{zad.6}\\\\\huge\boxed{a)}[/tex]
[tex]\sqrt{18} =\sqrt{9\cdot 2} =3\sqrt{2} \\\\\sqrt{2} \approx 1,41\\\\\huge\boxed{\sqrt{18} -3 > 0,~~5-3\sqrt{2} > 0,~~1-\sqrt{2} < 0}\\\\\dfrac{\mid \sqrt{18} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid = \dfrac{\mid 3\sqrt{2} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid =\dfrac{3\sqrt{2} -3}{-( 1-\sqrt{2}) } -2(5-3\sqrt{2} )=\dfrac{3\sqrt{2} -3}{\sqrt{2} -1} \cdot \dfrac{ \sqrt{2} +1}{\sqrt{2} +1 } -10+6\sqrt{2} =\dfrac{6+3\sqrt{2} -3\sqrt{2} -3}{(\sqrt{2})^{2} -1^{2}} -10+6\sqrt{2} =[/tex]
[tex]=\dfrac{3}{2-1} -10+6\sqrt{2} =3-10+6\sqrt{2}=\boxed{6\sqrt{2}-7}[/tex]
[tex]\huge\boxed{b)}[/tex]
[tex]\sqrt{3} \approx 1,73\\\\\huge\boxed{2-\sqrt{3} > 0,~~~~\sqrt{3} -2 < 0,~~~~2\sqrt{3} -5 < 0}[/tex]
[tex]\dfrac{\mid 2-\sqrt{3} \mid \cdot \mid \sqrt{3}-2 \mid }{ \mid 2\sqrt{3}-5 \mid} =\dfrac{( 2-\sqrt{3} ) \cdot -( \sqrt{3}-2 ) }{ -(2\sqrt{3}-5 )} =\dfrac{( 2-\sqrt{3} ) \cdot (2-\sqrt{3} ) }{ 5-2\sqrt{3}} =\dfrac{( 2-\sqrt{3} )^{2} }{ 5-2\sqrt{3}} \cdot \dfrac{ 5+2\sqrt{3}}{ 5+2\sqrt{3}} =\dfrac{( 7-4\sqrt{3} ) \cdot ( 5+2\sqrt{3}) }{ 5^{2}-(2\sqrt{3})^{2}}=\dfrac{35+14\sqrt{3} -20\sqrt{3} -24}{25-12} =\boxed{\dfrac{11-6\sqrt{3} }{13} }[/tex]
[tex]\huge\boxed{c)}[/tex]
[tex]\\\\(\sqrt{7} -2)\cdot (\sqrt{7} +2)=(\sqrt{7} )^{2} -2^{2} =7-4=3\\\\\\\sqrt{3} \approx 1,73\\\\(1-\sqrt{3} )\cdot (\sqrt{3} -1)=\sqrt{3} -1-3+\sqrt{3} =2\sqrt{3} -4\\\\\boxed{(\sqrt{7} -2)\cdot (\sqrt{7} +2) > 0,~~~~(1-\sqrt{3} )\cdot (\sqrt{3} -1) < 0,~~~~3\sqrt{3} -5 > 0}\\\\\\\dfrac{ \mid (\sqrt{7} -2)\cdot (\sqrt{7} +2)\mid}{\mid (1-\sqrt{3} )\cdot (\sqrt{3} -1)\mid } -\dfrac{\mid 3\sqrt{3} -5\mid }{2} =\\\\[/tex]
[tex]=\dfrac{ \mid 3\mid}{\mid (2\sqrt{3} -4)\mid } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{ 3}{-(2\sqrt{3} -4) } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{ 3}{4-2\sqrt{3} } \cdot \dfrac{ 4+2\sqrt{3} }{ 4+2\sqrt{3} } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{12+6\sqrt{3} }{16-12} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{12+6\sqrt{3} }{4} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{6+3\sqrt{3} }{2} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{6+3\sqrt{3}-(3\sqrt{3} -5) }{2} =\dfrac{6+3\sqrt{3}-3\sqrt{3} +5 }{2} =\boxed{5\dfrac{1}{2}}[/tex]
[tex]\huge\boxed{d)}[/tex]
[tex]\sqrt{6} \approx 2,45\\\\-(3-\sqrt{6} )\cdot (3-\sqrt{6} )=-(3-\sqrt{6} )^{2} =6\sqrt{6} -15\\\\\boxed{(3-\sqrt{6} )\cdot (\sqrt{6}-3 ) < 0,~~~~6\sqrt{6} -11 > 0,~~~~\sqrt{6} -2 > 0,~~~~2+\sqrt{6} > 0}\\\\\dfrac{\mid (3-\sqrt{6} )\cdot (3-\sqrt{6} )\mid }{\mid \sqrt{6} -2\mid \cdot \mid 2+\sqrt{6}\mid } +\dfrac{1}{2} \mid 6\sqrt{6} -11\mid =\dfrac{-(6\sqrt{6} -15) }{( \sqrt{6} -2)\cdot ( 2+\sqrt{6}) } +\dfrac{1}{2} (6\sqrt{6} -11)=[/tex]
[tex]=\dfrac{15-6\sqrt{6} }{( \sqrt{6} -2)\cdot ( \sqrt{6}+2) } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{( \sqrt{6})^{2} -2^{2} } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{6-4 } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{2 } +3\sqrt{6} -5\dfrac{1}{2} =7\dfrac{1}{2} -3\sqrt{3} +3\sqrt{6} -5\dfrac{1}{2} =\boxed{2}[/tex]
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Odpowiedź:
[tex]\huge\boxed{a)~~\dfrac{\mid \sqrt{18} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid = 6\sqrt{2} -7}[/tex]
[tex]\huge\boxed{b)~~\dfrac{\mid 2-\sqrt{3} \mid \cdot \mid \sqrt{3}-2 \mid }{ \mid 2\sqrt{3}-5 \mid} =\dfrac{11-6\sqrt{3} }{13} }[/tex]
[tex]\huge\boxed{c)~~\dfrac{ \mid (\sqrt{7} -2)\cdot (\sqrt{7} +2)\mid}{\mid (1-\sqrt{3} )\cdot (\sqrt{3} -1)\mid } -\dfrac{\mid 3\sqrt{3} -5\mid }{2} =5\frac{1}{2} }[/tex]
[tex]\huge\boxed{d)~~\dfrac{\mid (3-\sqrt{6} )\cdot (3-\sqrt{6} )\mid }{\mid \sqrt{6} -2\mid \cdot \mid 2-\sqrt{6}\mid } +\dfrac{1}{2} \mid 6\sqrt{6} -11\mid =2}[/tex]
Szczegółowe wyjaśnienie:
Wartość bezwzględna:[tex]\huge\boxed{\mid x\mid = \left \{ {{x~~dla~~x\geq 0} \atop {-x~~dla~~x < 0}} \right\} }[/tex]
Korzystamy ze wzorów skróconego mnożenia:
Obliczamy wartość wyrażeń:
[tex]\huge\boxed{zad.6}\\\\\huge\boxed{a)}[/tex]
[tex]\sqrt{18} =\sqrt{9\cdot 2} =3\sqrt{2} \\\\\sqrt{2} \approx 1,41\\\\\huge\boxed{\sqrt{18} -3 > 0,~~5-3\sqrt{2} > 0,~~1-\sqrt{2} < 0}\\\\\dfrac{\mid \sqrt{18} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid = \dfrac{\mid 3\sqrt{2} -3\mid}{\mid 1-\sqrt{2}\mid } -2\mid 5-3\sqrt{2} \mid =\dfrac{3\sqrt{2} -3}{-( 1-\sqrt{2}) } -2(5-3\sqrt{2} )=\dfrac{3\sqrt{2} -3}{\sqrt{2} -1} \cdot \dfrac{ \sqrt{2} +1}{\sqrt{2} +1 } -10+6\sqrt{2} =\dfrac{6+3\sqrt{2} -3\sqrt{2} -3}{(\sqrt{2})^{2} -1^{2}} -10+6\sqrt{2} =[/tex]
[tex]=\dfrac{3}{2-1} -10+6\sqrt{2} =3-10+6\sqrt{2}=\boxed{6\sqrt{2}-7}[/tex]
[tex]\huge\boxed{b)}[/tex]
[tex]\sqrt{3} \approx 1,73\\\\\huge\boxed{2-\sqrt{3} > 0,~~~~\sqrt{3} -2 < 0,~~~~2\sqrt{3} -5 < 0}[/tex]
[tex]\dfrac{\mid 2-\sqrt{3} \mid \cdot \mid \sqrt{3}-2 \mid }{ \mid 2\sqrt{3}-5 \mid} =\dfrac{( 2-\sqrt{3} ) \cdot -( \sqrt{3}-2 ) }{ -(2\sqrt{3}-5 )} =\dfrac{( 2-\sqrt{3} ) \cdot (2-\sqrt{3} ) }{ 5-2\sqrt{3}} =\dfrac{( 2-\sqrt{3} )^{2} }{ 5-2\sqrt{3}} \cdot \dfrac{ 5+2\sqrt{3}}{ 5+2\sqrt{3}} =\dfrac{( 7-4\sqrt{3} ) \cdot ( 5+2\sqrt{3}) }{ 5^{2}-(2\sqrt{3})^{2}}=\dfrac{35+14\sqrt{3} -20\sqrt{3} -24}{25-12} =\boxed{\dfrac{11-6\sqrt{3} }{13} }[/tex]
[tex]\huge\boxed{c)}[/tex]
[tex]\\\\(\sqrt{7} -2)\cdot (\sqrt{7} +2)=(\sqrt{7} )^{2} -2^{2} =7-4=3\\\\\\\sqrt{3} \approx 1,73\\\\(1-\sqrt{3} )\cdot (\sqrt{3} -1)=\sqrt{3} -1-3+\sqrt{3} =2\sqrt{3} -4\\\\\boxed{(\sqrt{7} -2)\cdot (\sqrt{7} +2) > 0,~~~~(1-\sqrt{3} )\cdot (\sqrt{3} -1) < 0,~~~~3\sqrt{3} -5 > 0}\\\\\\\dfrac{ \mid (\sqrt{7} -2)\cdot (\sqrt{7} +2)\mid}{\mid (1-\sqrt{3} )\cdot (\sqrt{3} -1)\mid } -\dfrac{\mid 3\sqrt{3} -5\mid }{2} =\\\\[/tex]
[tex]=\dfrac{ \mid 3\mid}{\mid (2\sqrt{3} -4)\mid } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{ 3}{-(2\sqrt{3} -4) } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{ 3}{4-2\sqrt{3} } \cdot \dfrac{ 4+2\sqrt{3} }{ 4+2\sqrt{3} } -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{12+6\sqrt{3} }{16-12} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{12+6\sqrt{3} }{4} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{6+3\sqrt{3} }{2} -\dfrac{ 3\sqrt{3} -5 }{2} =\dfrac{6+3\sqrt{3}-(3\sqrt{3} -5) }{2} =\dfrac{6+3\sqrt{3}-3\sqrt{3} +5 }{2} =\boxed{5\dfrac{1}{2}}[/tex]
[tex]\huge\boxed{d)}[/tex]
[tex]\sqrt{6} \approx 2,45\\\\-(3-\sqrt{6} )\cdot (3-\sqrt{6} )=-(3-\sqrt{6} )^{2} =6\sqrt{6} -15\\\\\boxed{(3-\sqrt{6} )\cdot (\sqrt{6}-3 ) < 0,~~~~6\sqrt{6} -11 > 0,~~~~\sqrt{6} -2 > 0,~~~~2+\sqrt{6} > 0}\\\\\dfrac{\mid (3-\sqrt{6} )\cdot (3-\sqrt{6} )\mid }{\mid \sqrt{6} -2\mid \cdot \mid 2+\sqrt{6}\mid } +\dfrac{1}{2} \mid 6\sqrt{6} -11\mid =\dfrac{-(6\sqrt{6} -15) }{( \sqrt{6} -2)\cdot ( 2+\sqrt{6}) } +\dfrac{1}{2} (6\sqrt{6} -11)=[/tex]
[tex]=\dfrac{15-6\sqrt{6} }{( \sqrt{6} -2)\cdot ( \sqrt{6}+2) } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{( \sqrt{6})^{2} -2^{2} } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{6-4 } +3\sqrt{6} -5\dfrac{1}{2} =\dfrac{15-6\sqrt{6} }{2 } +3\sqrt{6} -5\dfrac{1}{2} =7\dfrac{1}{2} -3\sqrt{3} +3\sqrt{6} -5\dfrac{1}{2} =\boxed{2}[/tex]