\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)

\(-18x^3+51x^2+9x-60=0\)

\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)

ĐKXĐ: z>0

pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)

<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)

<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)

<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)

<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)

<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)

<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)

vậy x=2

Một bài làm rất hay !

hihi thank ad vân ..

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

mọi người ưi giúp tui giải câu a thui nha tui giải đc câu b ròi làm ơn nhanh giúp thanks nhìu nhìu

\(x+y+z=2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032\)

<=>\(\left(x-29\right)-2\sqrt{x-29\cdot}+1+\left(y-6\right)-4\sqrt{y-6}+4+\left(z-2011\right)-6\sqrt{z-2011}+9=0\)

<=>\(\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)

cho 3 cái =0 là ra 

nhân 2 lên rồi rút về hằng đẳng thức là xong bạn ak cần mk giải ra ko

pt <=> \(2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032=x+y+z\)

 <=> \(x-29-2\sqrt{x-29}+1+y-6-4\sqrt{y-6}+4+z-2011-6\sqrt{z-2011}+9-2032=0\)

== đề sai à 

\(\frac{3x+6}{\sqrt{3x+7}+1}-\frac{x}{\sqrt{x+1}-1}=0\)

\(pt\Leftrightarrow\frac{3x+6}{\sqrt{3x+7}+1}-\left(\frac{1}{2}x+\frac{3}{2}\right)-\left(\frac{x}{\sqrt{x+1}-1}-\left(\frac{1}{2}x+\frac{3}{2}\right)\right)=0\)

\(\Leftrightarrow\frac{\left(\frac{3x+6}{\sqrt{3x+7}+1}\right)^2-\left(\frac{1}{2}x+\frac{3}{2}\right)^2}{\frac{3x+6}{\sqrt{3x+7}+1}+\frac{1}{2}x+\frac{3}{2}}-\frac{\left(\frac{x}{\sqrt{x+1}-1}\right)^2-\left(\frac{1}{2}x+\frac{3}{2}\right)^2}{\frac{x}{\sqrt{x+1}-1}+\frac{1}{2}x+\frac{3}{2}}=0\)

OK làm nốt nhé :VVV

đúng gì, làm xong vẫn trở lại để bài