Cho a,b,c > 0 thỏa mãn ab + bc + ca = 1.
Tính A = $\sqrt[]{(a^2+1)(b^2+1)(c^2+1)}$
Cho a,b,c > 0 thỏa mãn ab + bc + ca = 1. Tính A = $\sqrt[]{(a^2+1)(b^2+1)(c^2+1)}$
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Cho a,b,c > 0 thỏa mãn ab + bc + ca = 1.
Tính A = $\sqrt[]{(a^2+1)(b^2+1)(c^2+1)}$
Jezebel
nếu `a=b=c`
`=>3a^2=3b^2=3c^2=1`
`=>a^2=b^2=c^2=1/3`
thay vào A ta có
`A=sqrt{(1/3+1)^3}`
`=sqrt{64/27}`
`=8/((3sqrt{3})`
`=(8sqrt{3})/9`
học tốt .-.
Bơ
Giả sử $a=b=c$, ta có:
$ab+bc+ca=1$
$↔ 3a^2=1$
$↔ a^2=\dfrac{1}{3}$
Khi đó:
$A=\sqrt[]{(a^2+1)(b^2+1)(c^2+1)}$
$=\sqrt[]{(a^2+1)^3}$
$=\sqrt[]{\Bigg(\dfrac{1}{3}+1\Bigg)^3}$
$=\dfrac{8\sqrt[]{3}}{9}$