Tham khảo
` M=\frac{2}{5.14}+\frac{2}{7.18}+\frac{2}{9.22}+\frac{2}{11.26}+\frac{2}{13.30}`
`⇒M=\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+\frac{1}{9}.\frac{1}{11}+\frac{1}{11}.\frac{1}{13}+\frac{1}{13}.\frac{1}{15}`
`⇒M=\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{13.15}`
`⇒M=\frac{1}{2}.(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15})`
`⇒M=\frac{1}{2}.(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{13}-\frac{1}{15})`
`⇒M=\frac{1}{2}.(\frac{1}{5}-\frac{1}{15})`
`⇒M=\frac{1}{2}.\frac{2}{15}=\frac{1}{15}`
Mà `\frac{3}{46}<\frac{3}{45}=\frac{1}{15}`
Do đó `M>\frac{3}{46}`
`2) \frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10`
`⇒\frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=10-1-2-3-4`
`⇒\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0`
`⇒(123-x)(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19})=0`
Vì `\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19} \ne0`
`⇔123-x=0`
`⇔x=123`
Vậy `x=123`
