8^x+8^y+8^z>4^(x+1)+4^(y+1)+4^(z+1) x+y+z=6 giai giup mik vs 28/07/2021 Bởi Faith 8^x+8^y+8^z>4^(x+1)+4^(y+1)+4^(z+1) x+y+z=6 giai giup mik vs
Đáp án: Giải thích các bước giải: Ta có: \(8^{x}+8^{y}+8^{z}>4^{x+1}+4^{y+1}+4^{x+1}\)\(8^{x}+8^{x}+8^{2}>3\sqrt[3]{8^{x}\cdot 8^{x}\cdot 8^{2}}=12\cdot 4^{x}\)\(8^{y}+8^{y}+8^{2}>3\sqrt[3]{8^{y}\cdot 8^{y}\cdot 8^{2}}=12\cdot 4^{y}\)\(8^{z}+8^{z}+8^{2}>3\sqrt[3]{8^{z}\cdot 8^{z}\cdot 8^{2}}=12\cdot 4^{z}\)\(8^{x}+8^{y}+8^{z}>3\sqrt[3]{8^{x}\cdot 8^{y}\cdot 8^{z}}=\sqrt[3]{8^{6}}=192\)Cộng các vế ta được:\(3(8^{x}+8^{y}+8^{z}+64)>3(4^{x+1}+4^{y+1}+4^{x+1}+64)\)hay \(8^{x}+8^{y}+8^{z}>4^{x+1}+4^{y+1}+4^{x+1}\) Bình luận
Đáp án:
Giải thích các bước giải:
Ta có: \(8^{x}+8^{y}+8^{z}>4^{x+1}+4^{y+1}+4^{x+1}\)
\(8^{x}+8^{x}+8^{2}>3\sqrt[3]{8^{x}\cdot 8^{x}\cdot 8^{2}}=12\cdot 4^{x}\)
\(8^{y}+8^{y}+8^{2}>3\sqrt[3]{8^{y}\cdot 8^{y}\cdot 8^{2}}=12\cdot 4^{y}\)
\(8^{z}+8^{z}+8^{2}>3\sqrt[3]{8^{z}\cdot 8^{z}\cdot 8^{2}}=12\cdot 4^{z}\)
\(8^{x}+8^{y}+8^{z}>3\sqrt[3]{8^{x}\cdot 8^{y}\cdot 8^{z}}=\sqrt[3]{8^{6}}=192\)
Cộng các vế ta được:
\(3(8^{x}+8^{y}+8^{z}+64)>3(4^{x+1}+4^{y+1}+4^{x+1}+64)\)
hay \(8^{x}+8^{y}+8^{z}>4^{x+1}+4^{y+1}+4^{x+1}\)