Question

1) 75²⁰ và 45¹⁰.5³⁰

2) \(\sqrt{40+2}và\sqrt{40}+\sqrt{2}\)

so sánh

\(\frac{x}{3}=\frac{y}{4}vàxy=12\)

tìm x;y

Answer 1

Ta có :

1) 45^10 . 5^30= (5.9)^10 . 5^30 = 5^10 . 5^30 . 9^10 = 5^40 . 3^20 = 25^20 . 3^20=75^20

2)\(\sqrt{40+2}=\sqrt{42}<\sqrt{49}=7=6+1=\sqrt{36}+\sqrt{1}<\sqrt{40}+\sqrt{2}\)

Vậy \(\sqrt{40+2}<\sqrt{40}+\sqrt{2}\)

3)\(Cho\frac{x}{3}=\frac{y}{4}=k\Rightarrow x=3k;y=4k\)

Ta lại có:

\(xy=12\Rightarrow3k.4k=12\)

\(12.k^2=12\Rightarrow k^2=1\Rightarrow k=1:-1\)

\(Vơik=1\Rightarrow x=1.3=3;y=1.4=4\)

\(k=-1\Rightarrow x=-1.3=-3;y=-1.4=-4\)