《2025年考点集训与满分备考九年级数学全一册下人教版》

1. 计算：$\sin^{2}60^{\circ}+\tan45^{\circ}\cos60^{\circ}$.

答案： [解析]原式=$(\frac{\sqrt{3}}{2})^{2}+1\times\frac{1}{2}=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$.

2. 计算：$\cos60^{\circ}-2\sin^{2}45^{\circ}+\frac{2}{3}\tan^{2}60^{\circ}-\sin30^{\circ}$.

答案： [解析]原式=$\frac{1}{2}-2\times(\frac{\sqrt{2}}{2})^{2}+\frac{2}{3}\times(\sqrt{3})^{2}-\frac{1}{2}=\frac{1}{2}-2\times\frac{1}{2}+\frac{2}{3}\times3-\frac{1}{2}=\frac{1}{2}-1 + 2-\frac{1}{2}=1$.

3. 计算：$\tan30^{\circ}\sin60^{\circ}-\cos^{2}30^{\circ}+\cos^{2}45^{\circ}\tan45^{\circ}$.

答案： [解析]原式=$\frac{\sqrt{3}}{3}\times\frac{\sqrt{3}}{2}-(\frac{\sqrt{3}}{2})^{2}+(\frac{\sqrt{2}}{2})^{2}\times1=\frac{1}{2}-\frac{3}{4}+\frac{1}{2}=\frac{1}{4}$.

4. 计算：$\frac{\tan45^{\circ}}{\tan60^{\circ}+2\sin45^{\circ}}-\cos30^{\circ}$.(结果保留根号)

答案： [解析]原式=$\frac{1}{\sqrt{3}+2\times\frac{\sqrt{2}}{2}}-\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}-\frac{\sqrt{3}}{2}=\sqrt{3}-\sqrt{2}-\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}-\sqrt{2}$.

5. 计算：$\frac{\sin45^{\circ}-\tan45^{\circ}}{\cos^{2}60^{\circ}}+2\cos30^{\circ}\sin60^{\circ}$.(结果保留根号)

答案： [解析]原式=$\frac{\frac{\sqrt{2}}{2}-1}{(\frac{1}{2})^{2}}+2\times\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}=2\sqrt{2}-4+\frac{3}{2}=2\sqrt{2}-\frac{5}{2}$.

6. 计算：$\sqrt{(4\cos60^{\circ}-\tan60^{\circ})(3\tan30^{\circ}+4\cos^{2}45^{\circ})}$.

答案： [解析]原式=$\sqrt{(4\times\frac{1}{2}-\sqrt{3})[3\times\frac{\sqrt{3}}{3}+4\times(\frac{\sqrt{2}}{2})^{2}]}=\sqrt{(2 - \sqrt{3})(\sqrt{3}+2)}=\sqrt{4 - 3}=1$.

7. 计算：$2\tan45^{\circ}-\frac{1}{\sin30^{\circ}}-2\sin^{2}60^{\circ}$.

答案： [解析]原式=$2\times1-\frac{1}{\frac{1}{2}}-2\times(\frac{\sqrt{3}}{2})^{2}=2 - 2-2\times\frac{3}{4}=2 - 2-\frac{3}{2}=-\frac{3}{2}$.

8. 求值：$\tan^{2}60^{\circ}-\frac{\cos45^{\circ}}{\tan45^{\circ}-\sin45^{\circ}}$.(结果保留根号)

答案： [解析]原式=$(\sqrt{3})^{2}-\frac{\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{2}}{2}}=3-\frac{\sqrt{2}}{2-\sqrt{2}}=3-(\sqrt{2}+1)=3-\sqrt{2}-1=2-\sqrt{2}$.

9. 计算：$\frac{\tan45^{\circ}}{\sin60^{\circ}\tan60^{\circ}}-\sqrt{(\sin30^{\circ}-1)^{2}}+2\cos^{2}45^{\circ}$.

答案： [解析]原式=$\frac{1}{\frac{\sqrt{3}}{2}\times\sqrt{3}}-\sqrt{(\frac{1}{2}-1)^{2}}+2\times(\frac{\sqrt{2}}{2})^{2}=\frac{2}{3}-\frac{1}{2}+1=\frac{7}{6}$.

10. 计算：$2^{-1}+\sqrt{3}\cos30^{\circ}+\vert -5\vert -(\pi - 2022)^{0}$.

答案： [解析]原式=$\frac{1}{2}+\sqrt{3}\times\frac{\sqrt{3}}{2}+5 - 1=\frac{1}{2}+\frac{3}{2}+5 - 1=6$.