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Embed code

<iframe id="LTTP"  width="80%"  onload='this.height=document.getElementById("LTTP").scrollWidth*0.75+"px"' frameborder="0"  src='https://www.desmos.com/calculator/pwowcxjsiw?embed' style="border: 1px solid #ccc"></iframe>    

Abstract pictures

<img class="desc_img" src="https://images.cnblogs.com/cnblogs_com/wanghai0666/1538550/o_Cnblogs_zytp_14.jpg">    

Editing code

<LT>例1</LT>    

Feed code

<div STYLE="page-break-after: always;"></div>    

Choice of code

<div  class="Grid"><div  class="Grid-cell">$A.\alpha内有无数条直线与\beta平行$</div><div  class="Grid-cell">$B.\alpha内有两条相交直线与\beta平行$</div></div><div  class="Grid"><div  class="Grid-cell">$C.\alpha,\beta平行于同一条直线$</div> <div  class="Grid-cell">$D.\alpha,\beta垂直于同一个平面$</div></div>    

Conic

$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1(a>b>0)$    

Piecewise functions

$\left\{\begin{array}{l}{x=1}\\{y=1}\\{z=3}\end{array}\right.$    


$f(x)=\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$    

Multiple-choice options

<div  class="XZXX"  >$A.0.7$  $B.0.6$  $C.0.4$  $D.0.3$</div>    

<div  class="XZXX"  >$A.f(\cfrac{3}{2})>f(2)>f(3)$  $B.f(3)>f(2)>f(\cfrac{3}{2})$  $C.f(\cfrac{3}{2})>f(3)>f(2)$  $D.f(3)>f(\cfrac{3}{2})>f(2)$</div>    

<div  class="XZXX"  >$A.a < b < c$  $B.a < c < b$  $C.b < c < a$  $D.c < a < b$</div>    

<div  class="XZXX"  >$A.(-\infty,0)$  $B.(0,+\infty)$  $C.(1,+\infty)$  $D.(-\infty,1)$</div>    

<div  class="XZXX"  >$A.[2,6]$  $B.[4,8]$  $C.[\sqrt{2},3\sqrt{2}]$  $D.[2\sqrt{2},3\sqrt{2}]$</div>    

<div  class="XZXX"  >$A.[2,6]$  $B.[4,8]$  $C.[\sqrt{2},3\sqrt{2}]$  $D.[2\sqrt{2},3\sqrt{2}]$</div>    

<div  class="XZXX"  >$A.f(\cfrac{3}{2})>f(2)>f(3)$  $B.f(3)>f(2)>f(\cfrac{3}{2})$  $C.f(\cfrac{3}{2})>f(3)>f(2)$  $D.f(3)>f(\cfrac{3}{2})>f(2)$</div>    

1、<a  href="https://www.cnblogs.com/wanghai0666/p/9613192.html  "  target="_blank">由抽象函数不等式求参数的取值范围</a>;    

<div  class="XZXX"  >$A.[1,+\infty)$  $B.(1,+\infty)$  $C.[2,+\infty)$  $D.(2,+\infty)$</div>    

<div  class="XZXX"  >$A.[\cfrac{1}{6},1]$  $B.[\cfrac{2}{13},1]$  $C.[\cfrac{1}{6},\cfrac{4}{13}]$  $D.[\cfrac{1}{6},2\sqrt{2}]$</div>    

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Origin www.cnblogs.com/wanghai0666/p/12169660.html