类型 | 示例 | 计算过程 | 结果 |
约分计算法 | \(\frac{2x^2}{4x}\) | 分子分母同时除以 \(2x\) | \(\frac{x}{2}\) |
整体通分法 | \(\frac{1}{a}+\frac{1}{b}\) | 通分得到 \(\frac{b+a}{ab}\) | \(\frac{a+b}{ab}\) |
换元通分法 | 已知 \(\frac{1}{a}+\frac{1}{b}=3\),求 \(\frac{2a+3ab+2b}{a-2ab+b}\) 的值 | 令 \(m = \frac{1}{a}+\frac{1}{b}\),则原式 \(=\frac{2(a+b)+3ab}{(a+b)-2ab}=\frac{2(\frac{1}{a}+\frac{1}{b})+3}{\frac{1}{a}+\frac{1}{b}-2}=\frac{2m + 3}{m - 2}\),将 \(m = 3\) 代入得 \(\frac{2×3+3}{3-2}=9\) | 9 |
顺次相加法 | \(\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}\) | \(\frac{1}{1×2}=1-\frac{1}{2}\),\(\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}\),\(\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}\),所以原式 \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\) | \(\frac{3}{4}\) |
裂项相消法 | \(\frac{1}{n(n+1)}\) | 根据公式 \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\) | \(\frac{1}{n}-\frac{1}{n+1}\) |
消元法 | 已知 \(\frac{1}{x}+\frac{1}{y}=5\),\(\frac{1}{xy}=6\),求 \(\frac{1}{x^2}+\frac{1}{y^2}\) 的值 | 由 \(\frac{1}{x}+\frac{1}{y}=5\) 可得 \(\frac{x+y}{xy}=5\),又因为 \(\frac{1}{xy}=6\),\(x+y = 30\),则 \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{x^2+y^2}{x^2y^2}=\frac{(x+y)^2-2xy}{(xy)^2}=\frac{30^2-2×\frac{1}{6}}{(\frac{1}{6})^2}=\frac{900-\frac{1}{3}}{\frac{1}{36}}=32820\) | 32820 |
倒数求值法 | 已知 \(\frac{x^2-y^2}{x} = 3\),求 \(\frac{y}{x}\) 的值 | 由 \(\frac{x^2-y^2}{x}=3\) 可得 \(\frac{(x-y)(x+y)}{x}=3\),即 \(x-y = 3\frac{x}{x+y}\),设 \(\frac{y}{x}=k\),则 \(x = \frac{y}{k}\),代入上式得 \(\frac{y}{k}-y = 3\frac{\frac{y}{k}}{\frac{y}{k}+y}\),化简可得 \(k=\frac{1}{2}\) 或 \(k=-3\)(舍去) | \(\frac{1}{2}\) |
整体代入法 | 已知 \(x+\frac{1}{x}=3\),求 \(\frac{x^4+x^2+1}{x^2}\) 的值 | 将所求式子化简为 \(\frac{x^4+x^2+1}{x^2}=x^2+\frac{1}{x^2}+1=(x+\frac{1}{x})^2-1\),把 \(x+\frac{1}{x}=3\) 代入可得 \((x+\frac{1}{x})^2-1 = 3^2-1 = 8\) | 8 |
(图片来源网络,侵删)