Cel putin 2 va rog frumos.
Anexe:
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albatran:
cam grele...cine iti da din astea? adica atunci cand ajungi la asa ceva se prsupuneca estideja bun; adioca astea nu suntniste exercitii didactice, sa te ajutesa intelegi ciniste exercitide "filtrat" nisite concurenti.....ex9 si 10 par mai abordabile...mai vbim peste 2 h ,dac nu rezolva nimeni pana atunci
Răspunsuri la întrebare
Răspuns de
1
Salut,
AL 9:

AL 10:
![\\\\x=\dfrac{1}{2\sqrt5-4}+\sqrt2=\dfrac{1}{\sqrt{20}-\sqrt{16}}+\sqrt2=\dfrac{\sqrt{20}+\sqrt{16}}{(\sqrt{20}+\sqrt{16})(\sqrt{20}-\sqrt{16})}+\sqrt2=\\\\\\=\dfrac{2\sqrt5+4}{20-16}+\sqrt2=\dfrac{2\sqrt5+4}{4}+\sqrt2=\dfrac{\sqrt5+2}2+\sqrt2=\dfrac{\sqrt5}2+1+\sqrt2.\\\\\sqrt5>\sqrt4\Rightarrow\sqrt5>2\Rightarrow\dfrac{\sqrt5}2>1,\ deci\ \left[\dfrac{\sqrt5}2\right]=1;\\\\\[x\]=\left[\dfrac{\sqrt5}2\right]+[1]+[\sqrt2]=1+1+1=3,\ deci\ [x]=3. \\\\x=\dfrac{1}{2\sqrt5-4}+\sqrt2=\dfrac{1}{\sqrt{20}-\sqrt{16}}+\sqrt2=\dfrac{\sqrt{20}+\sqrt{16}}{(\sqrt{20}+\sqrt{16})(\sqrt{20}-\sqrt{16})}+\sqrt2=\\\\\\=\dfrac{2\sqrt5+4}{20-16}+\sqrt2=\dfrac{2\sqrt5+4}{4}+\sqrt2=\dfrac{\sqrt5+2}2+\sqrt2=\dfrac{\sqrt5}2+1+\sqrt2.\\\\\sqrt5>\sqrt4\Rightarrow\sqrt5>2\Rightarrow\dfrac{\sqrt5}2>1,\ deci\ \left[\dfrac{\sqrt5}2\right]=1;\\\\\[x\]=\left[\dfrac{\sqrt5}2\right]+[1]+[\sqrt2]=1+1+1=3,\ deci\ [x]=3.](https://tex.z-dn.net/?f=%5C%5C%5C%5Cx%3D%5Cdfrac%7B1%7D%7B2%5Csqrt5-4%7D%2B%5Csqrt2%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7B20%7D-%5Csqrt%7B16%7D%7D%2B%5Csqrt2%3D%5Cdfrac%7B%5Csqrt%7B20%7D%2B%5Csqrt%7B16%7D%7D%7B%28%5Csqrt%7B20%7D%2B%5Csqrt%7B16%7D%29%28%5Csqrt%7B20%7D-%5Csqrt%7B16%7D%29%7D%2B%5Csqrt2%3D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B2%5Csqrt5%2B4%7D%7B20-16%7D%2B%5Csqrt2%3D%5Cdfrac%7B2%5Csqrt5%2B4%7D%7B4%7D%2B%5Csqrt2%3D%5Cdfrac%7B%5Csqrt5%2B2%7D2%2B%5Csqrt2%3D%5Cdfrac%7B%5Csqrt5%7D2%2B1%2B%5Csqrt2.%5C%5C%5C%5C%5Csqrt5%26gt%3B%5Csqrt4%5CRightarrow%5Csqrt5%26gt%3B2%5CRightarrow%5Cdfrac%7B%5Csqrt5%7D2%26gt%3B1%2C%5C+deci%5C+%5Cleft%5B%5Cdfrac%7B%5Csqrt5%7D2%5Cright%5D%3D1%3B%5C%5C%5C%5C%5C%5Bx%5C%5D%3D%5Cleft%5B%5Cdfrac%7B%5Csqrt5%7D2%5Cright%5D%2B%5B1%5D%2B%5B%5Csqrt2%5D%3D1%2B1%2B1%3D3%2C%5C+deci%5C+%5Bx%5D%3D3.)
Green eyes.
AL 9:
AL 10:
Green eyes.
Răspuns de
1
11)
[tex]\it \left[\dfrac{x-2}{3}\right]=\dfrac{2x-4}{5},\ \ x\geq2 \\\;\\ \\\;\\ \underline{\mathcal{R}:}[/tex]
[tex]\it Fie\ k\in\mathbb{Z}. \\\;\\ \dfrac{2x-4}{5}=k \Rightarrow x=\dfrac{5k+4}{2} \Rightarrow x=\dfrac{5k}{2} +2\ \ \ (1) \\\;\\ \\\;\\ x\geq2 \Rightarrow \dfrac{5k}{2} +2 \geq2 \Rightarrow \dfrac{5k}{2} \geq0 \Rightarrow k\geq0 \ \ \ (2)[/tex]
Ecuația devine:
[tex]\it \left[\dfrac{\dfrac{5k}{2}+2-2}{3}\right]= k \Rightarrow \left[\dfrac{5k}{6}\right]= k \Rightarrow k\leq\dfrac{5k}{6}\ \textless \ k+1 \Rightarrow \\\;\\ \\\;\\ \Rightarrow 6k\leq5k\ \textless \ 6k+6 [/tex]
Din dubla inecuație, ținând seama de relația (2), rezultă k = 0, pe care îl
substituim în relația (1) și obținem x = 2.
Deci, S = {2}.
[tex]\it \left[\dfrac{x-2}{3}\right]=\dfrac{2x-4}{5},\ \ x\geq2 \\\;\\ \\\;\\ \underline{\mathcal{R}:}[/tex]
[tex]\it Fie\ k\in\mathbb{Z}. \\\;\\ \dfrac{2x-4}{5}=k \Rightarrow x=\dfrac{5k+4}{2} \Rightarrow x=\dfrac{5k}{2} +2\ \ \ (1) \\\;\\ \\\;\\ x\geq2 \Rightarrow \dfrac{5k}{2} +2 \geq2 \Rightarrow \dfrac{5k}{2} \geq0 \Rightarrow k\geq0 \ \ \ (2)[/tex]
Ecuația devine:
[tex]\it \left[\dfrac{\dfrac{5k}{2}+2-2}{3}\right]= k \Rightarrow \left[\dfrac{5k}{6}\right]= k \Rightarrow k\leq\dfrac{5k}{6}\ \textless \ k+1 \Rightarrow \\\;\\ \\\;\\ \Rightarrow 6k\leq5k\ \textless \ 6k+6 [/tex]
Din dubla inecuație, ținând seama de relația (2), rezultă k = 0, pe care îl
substituim în relația (1) și obținem x = 2.
Deci, S = {2}.
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