第121页

信息发布者：
＞
解：​$(2)\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+···+\frac {1}{\sqrt {2025}+\sqrt {2024}}$​
​$=\sqrt {2}-1+\sqrt {3}-\sqrt {2}+\sqrt {4}-\sqrt {3}+···+\sqrt {2025}-\sqrt {2024}$​
​$=\sqrt {2025}-1 $​
​$= 45 - 1$​
​$ = 44$​
​$(3)(\sqrt {9+x})^2-(\sqrt {3+x})^2=(\sqrt {9+x}+\sqrt {3+x})(\sqrt {9+x}-\sqrt {3+x})=9+x - 3 - x = 6$​
∵​$\sqrt {9+x}+\sqrt {3+x}=3$​
∴​$\sqrt {9+x}-\sqrt {3+x}=2$​
​$(4)$​∵​$(x+\sqrt {x^2+2025})(y+\sqrt {y^2+2025})=2025$​
∴​$x+\sqrt {x^2+2025}=\frac {2025}{y+\sqrt {y^2+2025}}=\sqrt {y^2+2025}-y$​
即​$x+\sqrt {x^2+2025}=\sqrt {y^2+2025}-y$​
同理可得，​$y+\sqrt {y^2+2025}=\sqrt {x^2+2025}-x$​
∴​$x + y+\sqrt {x^2+2025}+\sqrt {y^2+2025}=\sqrt {x^2+2025}+\sqrt {y^2+2025}-x - y$​
∴​$2(x + y)=0$​
则​$x + y = 0$​
上一页