SketchMath 數學團隊 SketchMath Maths Team ｜ 更新 Updated 2026-10-20
恆等式（恒等式）對變數的所有值都成立（記作 ≡），方程則只對某些值成立。常用恆等式：(a ± b)² = a² ± 2ab + b²、(a + b)(a − b) = a² − b²、a³ ± b³ = (a ± b)(a² ∓ ab + b²)。要求未知數，便比較同類項的係數。An identity is true for every value of the variable (written ≡), unlike an equation, which holds only for certain values. Key identities: (a ± b)² = a² ± 2ab + b², (a + b)(a − b) = a² − b², a³ ± b³ = (a ± b)(a² ∓ ab + b²). To find unknowns, compare coefficients of like terms.
(a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b²a² − b² = (a + b)(a − b)a³ + b³ = (a + b)(a² − ab + b²), a³ − b³ = (a − b)(a² + ab + b²)求 A、B、C，使 3x² + 5x − 2 ≡ A(x + 1)² + B(x + 1) + C。Find A, B and C such that 3x² + 5x − 2 ≡ A(x + 1)² + B(x + 1) + C.
Ax² + (2A + B)x + (A + B + C)Expand the right side: Ax² + (2A + B)x + (A + B + C)x²：A = 3。x：2A + B = 5，所以 B = −1。x²: A = 3. x: 2A + B = 5, so B = −1.A + B + C = −2，所以 C = −4Constant: A + B + C = −2, so C = −4答案： Answer: A = 3, B = −1, C = −4
證明 (x + 3)² − (x − 2)² ≡ 10x + 5。Show that (x + 3)² − (x − 2)² ≡ 10x + 5.
(x² + 6x + 9) − (x² − 4x + 4)= 10x + 5答案： Answer: 左方 = 右方，所以是恆等式。Left side = right side, so it is an identity.
不用計數機，求 998²。Evaluate 998² without a calculator.
(1000 − 2)² = 1000² − 2(1000)(2) + 2²1 000 000 − 4000 + 4答案： Answer: 996 004
(a + b)² ≠ a² + b²，必須加上 2ab。(a + b)² ≠ a² + b². Always include 2ab.溫習方法：點擊以下任何題目，即時在坐標網格載入並核對計算步驟。How to study: click any question to load it on the coordinate grid and check your steps.
SketchMath 題庫內的 DSE 風格練習，選擇題及長題目均附解答。 DSE-style practice from the SketchMath question bank, with solutions for every MC and long question.