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電子課本網(wǎng) › 第116頁

第116頁

信息發(fā)布者：
???$ 5\sqrt {\frac {1}{6}}$???
???$ 6\sqrt {\frac {1}{7}}$???
解：???$(2)$???原式???$=\sqrt {\frac {13×15+1}{15}}=\sqrt {\frac {142}{15}}=14\sqrt {\frac {1}{15}}$???
???$ (3)\sqrt {n+\frac {1}{n+2}}=(n+1)\sqrt {\frac {1}{n+2}}(n≥1)$??
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2
2
解：原式
???$={[(\sqrt {6}-\sqrt {5})×(\sqrt {6}+\sqrt {5})]}^{2021}$???
???$ ×(\sqrt {6}+\sqrt {5})$???
???$ =1×(\sqrt {6}+\sqrt {5})$???
???$ =\sqrt {6}+\sqrt {5}$???
解：原式???$=(1+\sqrt {3}-\sqrt {2})2$???
???$-(1+\sqrt {3}+\sqrt {2})2$???
???$ =-2\sqrt {2}×(2+2\sqrt {3})$???
???$ =-4\sqrt {2}-4\sqrt {6}$???
解：???$ (1)$???若???$a$???有意義，則???$8-x≥0，$??????$x≤8$???
若???$b$???有意義，則???$3x+4≥0，$??????$x≥-\frac {4}{3}$???
若???$c$???有意義，則???$x+2≥0，$??????$x≥-2$???
當???$-\frac {4}{3}≤x≤8$???時，???$a、$??????$b、$??????$c$???都有意義
???$ (2)$???若???$a、$??????$b、$??????$c$???為直角三角形的三邊，則???$-\frac {4}{3}< x< 8$???
???$ ①a2+b2= c2$???時，???$(8-x)+ (3x+4)=x+2$???
???$ x=-10，$???不滿足???$-\frac {4}{3}<x<8$???
故此時不成立
???$ ②a2+c2= b2$???時，???$(8-x)+(x+2)= 3x+4$???
???$ x=2，$???滿足???$-\frac {4}{3}<x< 8$???
???$ ③c2+b2=a2$???時，???$(3x+4)+(x+2)=8-x$???
???$ x=\frac {2}{5}，$???滿足???$-\frac {4}{3}<x<8$???
綜上所述：當???$x= 2$???或???$\frac {2}{5}$???時，???$a、$??????$b、$??????$c$???為直角三角形的三邊。

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??$2+\sqrt 3$??
??$\text {解: (3) 原式 }=\sqrt{(\sqrt{2})^{2}-2 \sqrt{2}+1}+ $??
??$\sqrt {(\sqrt {3})^2-2 ×\sqrt {3} ×\sqrt {2}+(\sqrt {2})^2}+ $??
??$\sqrt {2^2-2 ×2 ×\sqrt {3}+(\sqrt {3})^2} +···+$??
??$\sqrt {(\sqrt {n+1})^2+2 \sqrt {n+1} · \sqrt {n}+(\sqrt {n})^2} $??
??$=\sqrt {(\sqrt {2}-1)^2}+\sqrt {(\sqrt {3}-\sqrt {2})^2}+\sqrt {(2-\sqrt {3})^2} $??
??$+···+\sqrt {(\sqrt {n+1}-\sqrt {n})^2} $??
??$=\sqrt {2}-1+\sqrt {3}-\sqrt {2}+2-\sqrt {3}+···+\sqrt {n+1}-\sqrt {n} $??
??$=\sqrt {n+1}-1 .$??
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