使用带图形的解法解决一元一次方程
几何画板及绘图内容加载中,请稍等……
这个范例演示了如何通过带图形的解法来解决一元一次方程。核心数学概念包括方程的解、等式平衡和图形表示。作图要点涉及使用GeoGebra命令如Shuffle、Element、Sequence等来构造方程和图形。适合小学和初中生使用,帮助他们理解方程的解法。
GeoGebra 命令及参数
Shuffle({1, 2, 3, 4}) → allProblTypeAndC, {1, 3, 2, 4}Element(allProblTypeAndC, Mod[countQuestion, Length[allProblTypeAndC]] + 1) → problTypeShuffle(2…10) → allA, {3, 8, 9, 6, 10, 2, 4, 5, 7}Element(allA, Mod[countQuestion, Length[allA]] + 1) → valAShuffle(Sequence[valA + 1, 21]) → allB, {15, 17, 18, 11, 6, 10, 7, 9, 16, 19, 21, 12, 13, 14, 8, 20, 5, 4}Element(allB, Mod[countQuestion, Length[allB]] + 1) → valBSequence(n, n, 2, Min[valB - 1, 10]) → dummyA1RemoveUndefined(Zip[If[GCD[valB, n] ≟ 1, n], n, dummyA1]) → allA1Element(allA1, Mod[countQuestion, Length[allA1]] + 1) → valA1Element(allC, Mod[countQuestion, Length[allC]] + 1) → valCIf(problType ≟ 1, savedValue ≟ valB - valA, problType ≟ 2, savedValue ≟ valB + valA, problType ≟ 3, savedValue ≟ valB / valA, problType ≟ 4, savedValue ≟ ((valC * valB)) / valA1) → correctForDisplayIf(problType ≟ 1, Distance[T1, T2] / valB, problType ≟ 2, Distance[T1, T2] / (valA + valB), problType ≟ 4, Distance[T1, T2] / valB, 1) → unitTextfield(input1) → InputBox1If(problType ≟ 1, input1 ≟ valB - valA, problType ≟ 2, input1 ≟ valB + valA, problType ≟ 3, input1 ≟ valB / valA, problType ≟ 4, input1 ≟ ((valC * valB)) / valA1) → correct1Polygon(Zip[PixelsToPoint[Corner[1], Corner[3], Corner[5], A], A, {(pixelsLeft, pixelsTop), (pixelsLeft, y(Corner[5]) - pixelsBottom), Corner[5] - (pixelsRight, pixelsBottom), (x(Corner[5]) - pixelsRight, pixelsTop)}]) → borderBoxCorner(5) → layoutAppSizeSegment(L1, L2) → rhsPolygon(T1, T2, T3, T4) → tape, t1, t2, t3, t4Polygon(T1, T11, T41, T4) → rectXType1, f, g, h, iPolygon(T11, T2, T3, T41) → rectConstType1, j, k, l, mDistance(T1, T2) → lConstTapePolygon(T1, T12, T42, T4) → rectBType2, n, t12, t42, pPolygon(T12, T2, T3, T42) → rectAType2, q, r, s, tSequence(Polygon[(i + x(T1), y(T1)), (i + x(T1) + lHorCutsType3, y(T1)), (i + x(T1) + lHorCutsType3, y(T3)), (i + x(T1), y(T3))], i, 0, lConstTape - lHorCutsType3, lHorCutsType3) → unitCutsType3Midpoint(T1, T4) → M1Midpoint(T2, T3) → M2If(¬showanswer, Sequence[M1 + (i / ((2 * valA)) * (M2 - M1)), i, 1, (2 * valA), 2], Sequence[M1 + (i / ((2 * valA)) * (M2 - M1)), i, 1, (2 * valA), 2]) → allMidpointsSequence(labelX3, n, 1, Length[allMidpoints]) → allLabelsX3Segment(LT1, LT2) → xtype4Sequence(Polygon[(i + x(T1), y(T1)), (i + x(T1) + lHorCutsType4, y(T1)), (i + x(T1) + lHorCutsType4, y(T3)), (i + x(T1), y(T3))], i, 0, lConstTape - lHorCutsType4, lHorCutsType4) → unitCutsType4Polygon(T14, T2, T3, (x(T14), y(T14) + hTape)) → rectRemoveType4, t14, c, d, eSegment(L1, L4) → rhsType4Denominator((valC * valB / valA1)) → denX4Sequence(Polygon[(i + x(T1), y(T1)), (i + x(T1) + lHorCutsType4, y(T1)), (i + x(T1) + lHorCutsType4, y(T3)), (i + x(T1), y(T3))], i, (lHorCutsType4 * valA1), lConstTape - lHorCutsType4, lHorCutsType4) → unitCutsType4GrayedSequence(Text[allLabelsX3(A), allMidpoints(A) + If[¬showsolution, ((-0.3 * unit), 0.5), ((-0.5 * unit), 1)], true, true], A, 1, Length[allMidpoints]) → allLabelsType3Segment((0, 6), T31) → xtype3Segment((0, 6), (x(T11), 6)) → xtype1If(showsolution, Midpoint[LT1, LT2] - (1.5, 0) + If[denX4 ≠ 1, (-0.5, 2.5), (-0.5, 1.5)], Midpoint[LT1, LT2] - (1.5, 0) + If[denX4 ≠ 1, (-0.5, 2.5), (-0.5, 1.5)]) → PX4
AI 绘图提示语参考
在画布上构造一个用于一元一次方程可视化的“胶带图”(tape diagram):先放置一条水平长矩形作为总尺,再在其上方标出等分段并形成左右对称的划分。把矩形分成两块代表等式两边的量,并在中间用等长刻度表示同一比例的拆分。生成两端标记点及相应的分割点,连接形成胶带边界与分隔线;同时在胶带下方或上方放置若干文字输入框区域,用于显示方程的未知量关系。最后在胶带中找到两组对应的中点,按比例沿胶带方向放置一排标注点与标签,以便表达解的求法与对照区域。
以上提示语为根据原始绘图命令整理而来,仅供参考。