{"version":"1.0","provider_name":"Studijni-svet.cz","provider_url":"https:\/\/studijni-svet.cz\/kurzy","author_name":"Studijni-svet.cz","author_url":"https:\/\/studijni-svet.cz\/kurzy\/author\/admin\/","title":"\u0158e\u0161en\u00fd p\u0159\u00edklad 1 | Studijni-svet.cz","type":"rich","width":600,"height":338,"html":"<blockquote class=\"wp-embedded-content\" data-secret=\"aAF2ioX8XC\"><a href=\"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/reseny-priklad-1-2\/\">\u0158e\u0161en\u00fd p\u0159\u00edklad 1<\/a><\/blockquote><iframe sandbox=\"allow-scripts\" security=\"restricted\" src=\"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/reseny-priklad-1-2\/embed\/#?secret=aAF2ioX8XC\" width=\"600\" height=\"338\" title=\"&#8222;\u0158e\u0161en\u00fd p\u0159\u00edklad 1&#8220; &#8212; Studijni-svet.cz\" data-secret=\"aAF2ioX8XC\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" class=\"wp-embedded-content\"><\/iframe><script>\n\/*! This file is auto-generated *\/\n!function(d,l){\"use strict\";l.querySelector&&d.addEventListener&&\"undefined\"!=typeof URL&&(d.wp=d.wp||{},d.wp.receiveEmbedMessage||(d.wp.receiveEmbedMessage=function(e){var t=e.data;if((t||t.secret||t.message||t.value)&&!\/[^a-zA-Z0-9]\/.test(t.secret)){for(var s,r,n,a=l.querySelectorAll('iframe[data-secret=\"'+t.secret+'\"]'),o=l.querySelectorAll('blockquote[data-secret=\"'+t.secret+'\"]'),c=new RegExp(\"^https?:$\",\"i\"),i=0;i<o.length;i++)o[i].style.display=\"none\";for(i=0;i<a.length;i++)s=a[i],e.source===s.contentWindow&&(s.removeAttribute(\"style\"),\"height\"===t.message?(1e3<(r=parseInt(t.value,10))?r=1e3:~~r<200&&(r=200),s.height=r):\"link\"===t.message&&(r=new URL(s.getAttribute(\"src\")),n=new URL(t.value),c.test(n.protocol))&&n.host===r.host&&l.activeElement===s&&(d.top.location.href=t.value))}},d.addEventListener(\"message\",d.wp.receiveEmbedMessage,!1),l.addEventListener(\"DOMContentLoaded\",function(){for(var e,t,s=l.querySelectorAll(\"iframe.wp-embedded-content\"),r=0;r<s.length;r++)(t=(e=s[r]).getAttribute(\"data-secret\"))||(t=Math.random().toString(36).substring(2,12),e.src+=\"#?secret=\"+t,e.setAttribute(\"data-secret\",t)),e.contentWindow.postMessage({message:\"ready\",secret:t},\"*\")},!1)))}(window,document);\n\/\/# sourceURL=https:\/\/studijni-svet.cz\/kurzy\/wp-includes\/js\/wp-embed.min.js\n<\/script>\n","description":"Zad\u00e1n\u00ed: \u0158e\u0161t\u011b v oboru R rovnici: $$sqrt{x+27}=2+sqrt{x-5}$$ \u0158e\u0161en\u00ed: Za\u010dneme ur\u010den\u00edm podm\u00ednky \u017ee x\u22655 a n\u00e1sledn\u011b celou rovnici umocn\u00edme $${(sqrt{x+27})}^{2}={(2+sqrt{x-5})}^{2}$$ Nesm\u00edme zapomenout, \u017ee v\u00fdraz na prav\u00e9 stran\u011b rovnice mus\u00edme upravit podle vzorce (a+b)\u00b2=a\u00b2+2ab+b\u00b2, kde a=2 a b=\u221ax-5 $$x+27={2}^{2}+4sqrt{x-5}+(x-5)$$ $$x+27=4+4sqrt{x-5}+x-5$$ $$28=4sqrt{x-5}\/:4$$ $$7=sqrt{x-5}\/\u00b2$$ $$49=x-5$$ $$x=54$$ Je\u0161t\u011b provedeme zkou\u0161ku, abychom si ov\u011b\u0159ili spr\u00e1vnost \u0159e\u0161en\u00ed $$L(54)=sqrt{54+27}=sqrt{81}=9$$ $$P(54)=2+sqrt{54-5}=2+sqrt{49}=2+7=9$$ $$L=P$$ K={54} &nbsp; ... Read more"}