25. (9分)材料阅读:对非负数x“四舍五入”到个位的值记为$[x]$,即:当n为非负整数时,如果$\begin{cases}x≥ n-\dfrac{1}{2}, \\ x ＜ n+\dfrac{1}{2}, \end{cases}$则$[x]=n$.如:$[0]=[0.48]=0$, $[0.64]=[1.493]=1$, $[2]=2$, $···$
解决下列问题:
(1) ①填空:$[π]=$;
②如果$[2x - 1]=3$,求x的取值范围;
(2)判断:$[x + y]=[x]+[y]$是否成立?若成立,请说明理由;若不成立,请举出反例;
(3)请求出满足$[x]=\dfrac{4}{3}x$的所有非负数x的值.

25. 【点拨】本题考查新定义,解一元一次不等式组.
【解析】(1)①由题意得$[π]=3$.故答案为3.
②
∵ $[2x-1]=3$,
∴ 2.5 ≤ 2x - 1 ＜ 3.5,解得1.75 ≤ x ＜ 2.25.
(2)$[x+y]=[x]+[y]$不成立.举出反例如下:
若x=1.6,y=1.7,则$[x+y]=[1.6+1.7]=[3.3]=3$,$[x]+[y]=[1.6]+[1.7]=2+2=4$,
∴ $[1.6+1.7]≠[1.6]+[1.7]$,
∴ $[x+y]=[x]+[y]$不成立.
(3)设$\frac{4}{3}x=k$(k为非负整数),
∴ $x=\frac{3}{4}k$,即$[\frac{3}{4}k]=k$,
∴ $k-\frac{1}{2}≤\frac{3}{4}k＜k+\frac{1}{2}$,
∴ -2 ＜ k ≤ 2.
∵ k为非负整数,
∴ k=0或1或2.
当k=0时,x=0;当k=1时,x=$\frac{3}{4}$;当k=2时,x=$\frac{3}{2}$.
综上所述,x的值为0或$\frac{3}{4}$或$\frac{3}{2}$.

26. (9 分)在$△ ABC$中,$∠ C=60°$, 将线段$AB$沿直线$BC$平移得到线段$DE$(点$D$与点$B$对应, 且不与点$B,C$重合),连接$AE$,$∠ AED$和$∠ ACD$的平分线所在直线相交于点$P$(点$P$不与点$C,E$重合).
(1)如图,$∠ B=40°$;
①依题意,用无刻度直尺和圆规补全图;
②求$∠ EPC$的度数.
(2)若$∠ B=α$, 请直接写出$∠ EPC$的度数.(用含$α$的式子表示)
备用图
·78·

26. 【点拨】本题考查平行线的性质和判定,尺规作图作角平分线,角平分线的概念.
【解析】(1)①如图1所示,

②如图2所示,过点P作MN//AE.

∵ 将线段AB沿直线BC平移得到线段DE,
∴ AB//DE,
∴ ∠EDC = ∠ABC = 40°.
∵ AE//BD,
∴ ∠AED = ∠EDC = 40°.
∵ ∠AED和∠ACD的平分线所在直线相交于点P,
∴ ∠AEP = $\frac{1}{2}$∠AED = 20°,∠BCP = $\frac{1}{2}$∠ACB = 30°.
∵ AE//MN//BC,
∴ ∠EPN = ∠AEP = 20°,∠CPN = ∠BCP = 30°,
∴ ∠EPC = ∠EPN + ∠CPN = 50°.
(2)①如图3所示,过点P作MN//AE.

∵ 将线段AB沿直线BC平移得到线段DE,
∴ AB//DE,
∴ ∠EDC = ∠ABC = α.
∵ AE//BD,
∴ ∠AED = ∠EDC = α.
∵ ∠AED和∠ACD的平分线所在直线相交于点P,
∴ ∠AEP = $\frac{1}{2}$∠AED = $\frac{1}{2}$α,∠BCP = $\frac{1}{2}$∠ACB = 30°.
∵ AE//MN//BC,
∴ ∠EPN = ∠AEP = $\frac{1}{2}$α,∠CPN = ∠BCP = 30°,
∴ ∠EPC = ∠EPN + ∠CPN = $\frac{1}{2}$α + 30°.
②如图4所示,过点P作M₁N₁//AE,交AB于点M₁,交DE于点N₁,

∵ ∠ACB = 60°,
∴ ∠ACD = 120°.
∵ 将线段AB沿直线BC平移得到线段DE,
∴ AB//DE,
∴ ∠EDC₁ = ∠ABC = α.
∵ AE//BD,
∴ ∠AED = ∠EDC₁ = α.
∵ ∠AED和∠ACD的平分线所在直线相交于点P,
∴ ∠AEP = $\frac{1}{2}$∠AED = $\frac{1}{2}$α,∠ACP = $\frac{1}{2}$∠ACD = 60°,
∴ ∠BCP = ∠ACB + ∠ACP = 120°.
∵ AE//M₁N₁//BC,
∴ ∠EPN₁ = ∠AEP = $\frac{1}{2}$α,∠CPN₁ = ∠BCP = 120°,
∴ ∠EPC = ∠EPN₁ + ∠CPN₁ = $\frac{1}{2}$α + 120°.
③如图5所示,过点P作M₂N₂//AE,交AC于点N₂,交DE于点M₂.

∵ 将线段AB沿直线BC平移得到线段DE,
∴ AB//DE,
∴ ∠EDC = ∠ABC = α.
∵ AE//BD,
∴ ∠AED = 180° - ∠EDC = 180° - α.
∵ ∠AED和∠ACD的平分线所在直线相交于点P,
∴ ∠AEP = $\frac{1}{2}$∠AED = 90° - $\frac{1}{2}$α,∠BCP = $\frac{1}{2}$∠ACD = 30°.
∵ AE//M₂N₂//BC,
∴ ∠EPN₂ = 180° - ∠AEP = 90° + $\frac{1}{2}$α,∠CPN₂ = ∠BCP = 30°,
∴ ∠EPC = ∠EPN₂ + ∠CPN₂ = $\frac{1}{2}$α + 120°.
④如图6所示,过点P作M₃N₃//AE,

∵ 将线段AB沿直线BC平移得到线段DE,
∴ AB//DE,
∴ ∠EDC = ∠ABC = α.
∵ AE//BD,
∴ ∠AED = 180° - ∠EDC = 180° - α,
∴ 360° - (180° - α) = 180° + α.
∵ ∠AED和∠ACD的平分线所在直线相交于点P,
∴ ∠AEP = $\frac{1}{2}$(180° + α) = 90° + $\frac{1}{2}$α,∠BCP = $\frac{1}{2}$∠ACD = 30°.
∵ AE//M₃N₃//BC,
∴ ∠EPN₃ = 180° - ∠AEP = 90° - $\frac{1}{2}$α,∠CPN₃ = ∠BCP = 30°,
∴ ∠EPC = ∠EPN₃ - ∠CPN₃ = 60° - $\frac{1}{2}$α.
综上所述,∠EPC的度数为$\frac{1}{2}α + 30°$或$\frac{1}{2}α + 120°$或$60° - \frac{1}{2}α$.