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Đáp án: $T<P$ 

Giải thích các bước giải:

Ta có:

$P=\dfrac{155}{248}+\dfrac{3+\dfrac3{2025}-45\%}{8+\dfrac8{2025}-1.2}$

$\to P=\dfrac{155}{248}+\dfrac{3+\dfrac3{2025}-\dfrac9{20}}{8+\dfrac8{2025}-\dfrac65}$

$\to P=\dfrac{155}{248}+\dfrac{3(1+\dfrac1{2025}-\dfrac3{20})}{8(1+\dfrac1{2025}-\dfrac3{20})}$

$\to P= \dfrac58+\dfrac38$

$\to P=1$

Đặt $14^{2024}=a, 15^{2025}=b, 16^{2026}=c$

$\to T=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$

$\to T<\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}$

$\to T<1$

$\to T<P$