So sánh : a) $\sqrt[]{12+\sqrt[]{12+\sqrt[]{12 +\sqrt[]{15}}} }$ và 4
b) $\sqrt[]{2}$ + $\sqrt[]{5}$ + $\sqrt[]{15}$ + $\sqrt[]{24}$ và 13
So sánh : a) $\sqrt[]{12+\sqrt[]{12+\sqrt[]{12 +\sqrt[]{15}}} }$ và 4 b) $\sqrt[]{2}$ + $\sqrt[]{5}$ + $\sqrt[]{15}$ + $\sqrt[]{2
By Alaia
`a)`
`\sqrt(15)<4`
`⇔\sqrt(15)+12<12+4=16`
`⇔\sqrt(\sqrt(15)+12)<4`
`⇔12+\sqrt(\sqrt(15)+12)<16`
`⇔\sqrt(12+\sqrt(\sqrt(15)+12))<4`
`⇔12+\sqrt(12+\sqrt(\sqrt(15)+12))<16`
`⇔\sqrt(12+\sqrt(12+\sqrt(\sqrt(15)+12)))<4`
`b)`
`\sqrt(15)<4`
`\sqrt(24)<5`
`\sqrt(2)<1,5`
`\sqrt(5)<2,5`
`⇒\sqrt(15)+\sqrt(24)+\sqrt(5)+\sqrt(2)<5+4+1,5+2,5=13`
`
a)Ta có $4^2=16\Rightarrow \sqrt{15}<4\\\Rightarrow 12+\sqrt{15}<16\Rightarrow \sqrt{12+\sqrt{15}}<4$
Tương tự như vậy suy ra
$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{15}}}}<4$
b)Ta có $\sqrt{15}<4;\ \sqrt{24}<5\\\Rightarrow \sqrt{15}+\sqrt{24}<9$
$1<\sqrt{2}<1,5;\ 2<\sqrt{5}<2,5\\\Rightarrow 3<\sqrt{2}+\sqrt{5}<4$
$\Rightarrow \sqrt{2}+\sqrt{5}+\sqrt{15}+\sqrt{24}<13$