Đặt \(\sqrt[3]{6-2\sqrt{7}}=a\), \(\sqrt[3]{6+2\sqrt{7}}=b\)

\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=12\\ab=2\end{matrix}\right.\)

\(x=\sqrt[3]{6-2\sqrt{7}}+\sqrt[3]{6+2\sqrt{7}}=a+b\)

\(\Rightarrow x^3=a^3+b^3+3ab\left(a+b\right)=12+3.2\left(a+b\right)=12+6x\)

\(\Rightarrow x^3-6x-12=0\)

\(Q=x^3-6x+17=\left(x^3-6x-12\right)+29=29\)

Bài 4:

a) Ta có: \(\widehat{yOz}+\widehat{xOy}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{yOz}=180^0-\widehat{xOy}=180^0-50^0=130^0\)

b) Ta có: \(\widehat{zOt}=\widehat{yOt}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}.130^0=65^0\)(do Ot là tia phân giác \(\widehat{yOz}\))

c) Ta có: \(\widehat{xOt}=\widehat{yOt}+\widehat{xOy}=65^0+50^0=115^0\)

Bài 5: 

 a) Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{xOz}=180^0-\widehat{xOy}=180^0-110^0=70^0\)

b) Ta có: \(\widehat{zOt}=\dfrac{1}{2}\widehat{xOz}=\dfrac{1}{2}.70^0=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))

c) Ta có: \(\widehat{xOt}=\widehat{zOt}=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))

Ta có: Ox//CD(gt)

\(\Rightarrow\widehat{OCD}+\widehat{COx}=180^0\)( 2 góc trong cùng phía)

\(\Rightarrow\widehat{COx}=180^0-\widehat{OCD}=180^0-120^0=60^0\)

\(\Rightarrow\widehat{ACx}=\widehat{AOC}-\widehat{COx}=110^0-60^0=50^0\)

Ta có: \(\widehat{ACx}+\widehat{OAB}=50^0+130^0=180^0\)

Mà 2 góc này là 2 góc trong cùng phía

=> AB//Cx//CD

\(\dfrac{2}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}-2}+\dfrac{12}{\sqrt{3}+3}=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{\sqrt{3}+2}{-1}+\dfrac{12\left(\sqrt{3}-3\right)}{-6}=\sqrt{3}+1+\sqrt{3}+2-2\sqrt{3}+6=9\)

e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\Rightarrow6,5-2x=\dfrac{13}{10}.\dfrac{5}{13}=\dfrac{1}{2}\Rightarrow2x=6,5-\dfrac{1}{2}=6\Rightarrow x=3\)f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=-\dfrac{1}{6}+\dfrac{3}{4}\Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)

g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\Rightarrow5x-15=6x+9\Rightarrow x=-24\) 

h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\Rightarrow\left(x-5\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)

a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)

b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)

c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)

d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)

Tóm tắt:

\(45km/h\rightarrow1h\)

\(60km/h\rightarrow?h\)

              Thời gian đi từ a đến b với vận tốc 60km/h là:

                         \(\dfrac{60.1}{45}=\dfrac{4}{3}\left(h\right)\)    

\(x^3+27=9-x^2\Leftrightarrow x^3+x^2+18=0\Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)\Leftrightarrow x+3=0\Leftrightarrow x=-3\)(do \(x^2-2x+6=\left(x-1\right)^2+5\ge5>0\))