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miss Akunina [59]
3 years ago
9

I need help. this is the problem x^2/3+10=7x^1/3

Mathematics
1 answer:
Arlecino [84]3 years ago
7 0

\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow  (x^{\frac{1}{3}})^2-7x^{\frac{1}
  {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0

\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5  \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125

S = {8; 125}


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