Matematică, întrebare adresată de Alesica08, 8 ani în urmă

Va rog repede, am nevoie de rezolvare​

Anexe:

Răspunsuri la întrebare

Răspuns de tcostel
0

 

\displaystyle\bf\\\textbf{Toate exercitiile se rezolva in multimea numerelor naturale)}\\a)\\x^2:\Big(3^2\Big)^2=2^2\\\\x^2:9^2=4\\\\x^2:81=4\\\\x^2=4\times81\\\\x^2=324\\\\x=\sqrt{324}\\\\\boxed{\bf~x=18}\\\\\\b)\\196:x^2=7^4:7^2\\\\196:x^2=7^{4-2}\\\\196:x^2=7^2\\\\196:x^2=49\\\\x^2=196:49\\\\x^2=4\\x=\sqrt{4}\\\\\boxed{\bf~x=2}

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c)\\\displaystyle\bf\\\textbf{Transformam numarul 111 din baza x in baza 10.}\\\\111_x = x^2+x^1+x^0=73_{10}\\\\\textbf{Rezolvam ecuatia:}\\\\x^2+x+1=73\\\\x^2+x+1=73\\\\x^2+x+1-73=0\\\\x^2+x-72=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1+4\cdot72}}{2}=\\\\=\frac{-1\pm\sqrt{1+288}}{2}=\frac{-1\pm\sqrt{289}}{2}=\frac{-1\pm17}{2}\\\\x_1=\frac{-1+17}{2}=\frac{16}{2}=\boxed{\bf8}\\\\x_2<0~~(solutie~eliminata)\\\\\boxed{\bf~x=8}\\\\\implies~\boxed{\bf111_8=73_{10}}

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\displaystyle\bf\\d)\\324:5^x=\Big(2\cdot3^2\Big)^2\\\\324:5^x=\Big(2\cdot9\Big)^2\\\\324:5^x=18^2\\\\324:5^x=324\\\\5^x=324:324\\\\5^x=1\\\\\boxed{\bf~x=0}

 

 

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