\(\dfrac{x}{3}=\dfrac{x}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{125}{62}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=28.2=56\end{matrix}\right.\)

áp dụng đẳng thức AM - GM cho 7 số :3 số \(a^7,3\) số \(b^7\) và số 1,ta có

\(3a^7+3b^7+1\ge7^7\sqrt{a^{21}.b^{21}1}=7a^7b^7\left(1\right)\)

tương tự

\(3a^7+3b^7+1\ge7b^3c^3\left(2\right)\);\(3c^7+3a^7+1\ge7c^3a^3\left(3\right)\)

công thức về các bất đẳng thức (1);(2);(3) ta được

\(6\left(a^7+b^7+c^7\right)+3\ge7\left(a^3b^3+b^3c^3+c^3a^3\right)\)

\(\Leftrightarrow6\left(a^7+b^7+c^7\right)+3\ge7.3\)

\(\Leftrightarrow a^7+b^7+c^7\ge3\left(đpcm\right)\)

\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)

\(\dfrac{5}{2\sqrt{5}}=\dfrac{\left(\sqrt{5}\right)^2}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{20}}{60}=\dfrac{2\sqrt{5}}{2.30}=\dfrac{\sqrt{5}}{30}\)

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\sqrt{y}\left(\sqrt{y}+b\right)}{\sqrt{y}.b}=\dfrac{\sqrt{y}+b}{b}\)

If you want to attend the course, you______pass the examinatin. It is a requiement.
A. must     B. should    C. can    D. could

\(SA\perp\left(ABCD\right)\)

\(\Rightarrow SA\perp BD\)

ABCD là hình thoi

\(\Rightarrow AC\perp BD\\ \Rightarrow BD\perp\left(SAC\right)\)

     \(BD\in\left(SBD\right)\Rightarrow\left(SBD\right)\perp\left(SAC\right)\)

a)\(\sqrt{0,36a^2}=0,6\left|a\right|=-0,6a\)

b)\(\sqrt{a^4\left(3-a\right)^2}=\sqrt{a^4}\sqrt{\left(3-a\right)^2}=\left|a^2\right|.\left|3-a\right|\)

vì \(a\ge0\) nên \(\left|a^2\right|=a^2\)

vì \(a\ge3nên\left|3-a\right|=a-3\)

vậy \(\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)

c)

\(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.6\left(1-a\right)^2}\\ =9.4\left|1-a\right|\)

vì a>1nên1-a<0d= do đó \(\left|1-a\right|=0-1\)

vậy \(\sqrt{27.48\left(1-a\right)^2}=36\left(a-1\right)\)

d)\(\dfrac{1}{a-b}:\sqrt{a^4\left(a-b\right)^2}=\dfrac{1}{a-b}:\left(\sqrt{10^4}\sqrt{\left(a-b\right)^2}\right)\)

\(=\dfrac{1}{a-b}:\left(a^2\left(a-b\right)\right)\)

gọi số cần tìm là \(\overline{ab}\)

theo bài ra ta có:\(\left\{{}\begin{matrix}a+b=5\\a^2+b^2=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a+b\right)^2-2ab=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\5^2-2ab=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=6\end{matrix}\right.\)

khi đó 2 số a,b là nghiệm P.trình

\(x^2-5x+6=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

vậy số tự nhiên cần tìm là 23 hoặc 32

\(\dfrac{x}{5}=\dfrac{2}{3}\\ \Rightarrow3x=2.5=10\\ \Rightarrow x=\dfrac{10}{3}\)