24-12-2022 | 652
Cho \(xyz = 2022\). Tính giá trị của biểu thức sau: \(Q = \frac{x}{{\frac{{xy}}{{2022}} + x + 1}} + \frac{y}{{yz + y + 2022}} + \frac{z}{{xz + z + 1}}\)
Bài giải
\(Q = \frac{x}{{\frac{{xy}}{{2022}} + x + 1}} + \frac{y}{{yz + y + 2022}} + \frac{z}{{xz + z + 1}}\)
\( = \frac{x}{{\frac{{xy}}{{xyz}} + x + 1}} + \frac{y}{{yz + y + xyz}} + \frac{z}{{xz + z + 1}}\) (thay \(2022=xyz\))
\(= \frac{x}{{\frac{1}{z} + x + 1}} + \frac{y}{{y(z + 1 + xz)}} + \frac{z}{{xz + z + 1}}\)
\(= \frac{x}{{\frac{{1 + xz + z}}{z}}} + \frac{y}{{y(xz + z + 1)}} + \frac{z}{{xz + z + 1}}\)
\(= \frac{{xz}}{{xz + z + 1}} + \frac{1}{{xz + z + 1}} + \frac{z}{{xz + z + 1}}\)
\(= \frac{{xz + 1 + z}}{{xz + z + 1}}\)
\(=1\)
