Odpowiedź:
a ) ( 5√3)² = 25*3 = 75 ( 6√2)² = 36*2 = 72
więc
5√3 > 6√2
-----------------
b) ( 7√2)² = 49*2 = 98 ( 4√6)² = 16*6 = 96
7√2 > 4√7
--------------------
c)
( [tex]\frac{3}{4} \sqrt{12} )^2 = \frac{9}{16} *12 = \frac{27}{4} = 6 \frac{3}{4}[/tex] ( [tex]\frac{2}{3} \sqrt{15})^2 = \frac{4}{9} *15 = \frac{20}{3} = 6 \frac{2}{3}[/tex]
[tex]\frac{3}{4} \sqrt{12} > \frac{2}{3} \sqrt{15}[/tex]
-----------------------
d )
( 2 [tex]\sqrt[3]{7} )^3 = 8*7 = 56[/tex] ( 3 [tex]\sqrt[3]{2} )^3 = 27*2 = 54[/tex]
2[tex]\sqrt[3]{7} > 3\sqrt[3]{2}[/tex]
Szczegółowe wyjaśnienie:
a) 5 pierwiastek 3
b) 7 pierwiastek 2
c) 3/4 pierwiastek 12
d) 2 pierwiastek(3) 7
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Odpowiedź:
a ) ( 5√3)² = 25*3 = 75 ( 6√2)² = 36*2 = 72
więc
5√3 > 6√2
-----------------
b) ( 7√2)² = 49*2 = 98 ( 4√6)² = 16*6 = 96
więc
7√2 > 4√7
--------------------
c)
( [tex]\frac{3}{4} \sqrt{12} )^2 = \frac{9}{16} *12 = \frac{27}{4} = 6 \frac{3}{4}[/tex] ( [tex]\frac{2}{3} \sqrt{15})^2 = \frac{4}{9} *15 = \frac{20}{3} = 6 \frac{2}{3}[/tex]
więc
[tex]\frac{3}{4} \sqrt{12} > \frac{2}{3} \sqrt{15}[/tex]
-----------------------
d )
( 2 [tex]\sqrt[3]{7} )^3 = 8*7 = 56[/tex] ( 3 [tex]\sqrt[3]{2} )^3 = 27*2 = 54[/tex]
więc
2[tex]\sqrt[3]{7} > 3\sqrt[3]{2}[/tex]
-----------------------
Szczegółowe wyjaśnienie:
a) 5 pierwiastek 3
b) 7 pierwiastek 2
c) 3/4 pierwiastek 12
d) 2 pierwiastek(3) 7