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5. (1)已知$(m + n)^2 = 11$，$mn = 2$，则$(m - n)^2$的值是
(2)已知$(x + y)^2 = 12$，$(x - y)^2 = 4$，则$x^2 + 3xy + y^2$的值是
(3)已知$a + b = 5$，$ab = -24$，则$a^2 + b^2$的值是
3
。(2)已知$(x + y)^2 = 12$，$(x - y)^2 = 4$，则$x^2 + 3xy + y^2$的值是
14
。(3)已知$a + b = 5$，$ab = -24$，则$a^2 + b^2$的值是
73
，$a - b$的值是±11
。
答案：
5.
(1)3
(2)14
(3)73 ±11
(1)3
(2)14
(3)73 ±11
6. 已知$x + y = 6$，$xy = 7$，求$(3x + y)^2 + (x + 3y)^2$的值。
答案：
6.解：原式$=10x^2 + 10y^2 + 12xy = 10(x + y)^2 - 8xy。$$\becausex + y = 6，$xy = 7，$\therefore$原式$=10×6^2 - 8×7=304。$
7. 已知非零实数$a$满足$a^2 - 3a + 1 = 0$，求$a^2 + \frac{1}{a^2}$和$a^4 + \frac{1}{a^4}$的值。
答案：
7.解：$\becausea\neq0$且$a^2 - 3a + 1 = 0，$$\therefore$两边同时除以a，得$a+\frac{1}{a}=3，$
$\therefore(a+\frac{1}{a})^2=3^2，$即$a^2 + 2+\frac{1}{a^2}=9，$$\thereforea^2+\frac{1}{a^2}=7，$
$\therefore(a^2+\frac{1}{a^2})^2=7^2，$即$a^4 + 2+\frac{1}{a^4}=49，$$\thereforea^4+\frac{1}{a^4}=47。$
$\therefore(a+\frac{1}{a})^2=3^2，$即$a^2 + 2+\frac{1}{a^2}=9，$$\thereforea^2+\frac{1}{a^2}=7，$
$\therefore(a^2+\frac{1}{a^2})^2=7^2，$即$a^4 + 2+\frac{1}{a^4}=49，$$\thereforea^4+\frac{1}{a^4}=47。$
8. 计算：$(1 - \frac{1}{2^2}) × (1 - \frac{1}{3^2}) × (1 - \frac{1}{4^2}) × \cdots × (1 - \frac{1}{2024^2}) × (1 - \frac{1}{2025^2})$。
答案：
8.解：原式$=(1+\frac{1}{2})×(1-\frac{1}{2})×(1+\frac{1}{3})×(1-\frac{1}{3})×$
$(1+\frac{1}{4})×(1-\frac{1}{4})×\cdots×(1+\frac{1}{2024})×(1-\frac{1}{2024})×$
$(1+\frac{1}{2025})×(1-\frac{1}{2025})=\frac{3}{2}×\frac{1}{2}×\frac{4}{3}×\frac{2}{3}×\frac{5}{4}×$
$\frac{3}{4}×\cdots×\frac{2025}{2024}×\frac{2023}{2024}×\frac{2026}{2025}×\frac{2024}{2025}=\frac{1}{2}×\frac{2026}{2025}=\frac{1013}{2025}$
$(1+\frac{1}{4})×(1-\frac{1}{4})×\cdots×(1+\frac{1}{2024})×(1-\frac{1}{2024})×$
$(1+\frac{1}{2025})×(1-\frac{1}{2025})=\frac{3}{2}×\frac{1}{2}×\frac{4}{3}×\frac{2}{3}×\frac{5}{4}×$
$\frac{3}{4}×\cdots×\frac{2025}{2024}×\frac{2023}{2024}×\frac{2026}{2025}×\frac{2024}{2025}=\frac{1}{2}×\frac{2026}{2025}=\frac{1013}{2025}$
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