Question

*Sa se determine termenul care il contine pe X in dezvoltarea :
(1/x² + ³√x)¹⁰. Vreau rezolvarea completa va rog.

Answer 1

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Answer 2

   
[tex]\displaystyle\\ \text{Avem binomul:}\\\\ \left(\frac{1}{x^2}+ \sqrt[\b3]{x}\right)^{10}\\\\ \text{Folosim formula (Binomul lui Newton):}\\ (a+b)^n = C_n^0a^n+C_n^1a^{n-1}b+C_n^2a^{n-2}b^2+\cdots +C_n^ka^{n-k}b^k+\cdots C_n^nb^n\\\\\\ [/tex]


[tex]\displaystyle\\ \left(\frac{1}{x^2}+ \sqrt[\b3]{x}\right)^{10}=\\\\ =C_{10}^0\cdot \left(\frac{1}{x^2}\right)^{10}+C_{10}^1\cdot \left(\frac{1}{x^2}\right)^9\cdot \sqrt[\b3]{x} +C_{10}^2\cdot \left(\frac{1}{x^2}\right)^8\cdot \Big(\sqrt[\b3]{x}\Big)^2+\\\\ +C_{10}^3\cdot \left(\frac{1}{x^2}\right)^7\cdot \Big(\sqrt[\b3]{x}\Big)^3+ C_{10}^4\cdot \left(\frac{1}{x^2}\right)^6\cdot \Big(\sqrt[\b3]{x}\Big)^4+\\ [/tex]


[tex]\displaystyle\\ +C_{10}^5\cdot \left(\frac{1}{x^2}\right)^5\cdot \Big(\sqrt[\b3]{x}\Big)^5+ +C_{10}^6\cdot \left(\frac{1}{x^2}\right)^4\cdot \Big(\sqrt[\b3]{x}\Big)^6+\\\\ +C_{10}^7\cdot \left(\frac{1}{x^2}\right)^3\cdot \Big(\sqrt[\b3]{x}\Big)^7+ C_{10}^8\cdot \left(\frac{1}{x^2}\right)^2\cdot \Big(\sqrt[\b3]{x}\Big)^8+\\\\ +C_{10}^9\cdot \left(\frac{1}{x^2}\right)\cdot \Big(\sqrt[\b3]{x}\Big)^9+ C_{10}^{10}\cdot \Big(\sqrt[\b3]{x}\Big)^{10}[/tex]


[tex]\displaystyle\\ \text{Termenul general este:}\\\\ C_{10}^k\cdot \left(\frac{1}{x^2}\right)^{10-k}\cdot \Big(\sqrt[\b3]{x}\Big)^k\\\\ \text{Scriem ecuatia:}\\\\ \left(\frac{1}{x^2}\right)^{10-k}\cdot \Big(\sqrt[\b3]{x}\Big)^k=x \\\\ \Big(x^{-2}\Big)^{10-k}\cdot \left(x^{ \dfrac{1}{3}}\right)^k=x\\\\\\ x^{-2(10-k)}\cdot x^{ \dfrac{k}{3}}=x\\\\ x^{-2(10-k) +\dfrac{k}{3}}=x\\\\ -2(10-k) +\dfrac{k}{3}=1~~\Big|\cdot 3\\\\ -6(10-k)+k=3\\\\ k-6(10-k)=3\\\\ k-60+6k=3\\\\ 7k=63\\\\ k= \frac{63}{7}=\boxed{\bf 9} [/tex]


[tex]\displaystyle\\ \Longrightarrow\text{Termenul este:}\\\\ \boxed{C_{10}^9\cdot \left(\frac{1}{x^2}\right)^{10-9}\cdot \Big(\sqrt[\b3]{x}\Big)^9} = 10\cdot \frac{1}{x^2}\cdot x^3=10\cdot \frac{x^3}{x^2}= \boxed{\boxed{10x}} [/tex]