Question

dau coroana!!
11 calc [x] = ? pt:
a) x= [√ 2019] × (2+√2) {-√2}
b) x=√n^2-2n , n>= 2
​c) x= (√3 +√7)^2

Answer 1

Răspuns:

Explicație pas cu pas:

a)

$x=[\sqrt{2019}]+(2+\sqrt{2})\{\--\sqrt{2}\}=\\44+(2+\sqrt{2})(-\sqrt{2}-[-\sqrt{2}])=\\44+(2+\sqrt{2})(-\sqrt{2}-(-2))=\\44+(2+\sqrt{2})(2-\sqrt{2})=\\44+2=46\\44^2\leq 2019\leq 45^2$

$[x]=46$

b)

$x=\sqrt{n^2-2n}=\sqrt{n^2-2n+1-1}=\sqrt{(n-1)^2-1}\\\sqrt{(n-2)^2}\leq \sqrt{(n-1)^2-1}\\n-2\leq \sqrt{(n-1)^2+1}$

$[x]=n-2$

c)

$x=(\sqrt3}+\sqrt{7})^2=3+2\sqrt{21}+7=10+2\sqrt{21}\\4^2\leq \sqrt{21}\leq 5^2\\32\leq 2\sqrt{21}\leq 50\\42\leq 10+2\sqrt{21}\leq 60$

$[x]=42$