T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(

\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)

\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)

\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)

\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)

\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

a); b) Do tích = 0 

=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)

=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)

a; *x-1=0 <=>x=1

    *2x+5=0 <=>x=-2,5

    *x²+2=0 <=> ko có x

b; tương tự a

a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left(x+1\right)\left(x+2\right)\left(x-4\right)\left(x-5\right)=-30\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^2-3x-4\right)\left(x^2-3x-5\right)=-30\)

Đặt x^2-3x=a

=>(a+1)(a-4)(a-5)=-30

=>\(\left(a^2-3a-4\right)\left(a-5\right)=-30\)

=>\(a^3-5a^2-3a^2+15a-4a+20+30=0\)

=>a^3-8a^2+11a+50=0

=>a=-1,77

=>x^2-3x=-1,77

=>x^2-3x+1,77=0

hay \(\left[{}\begin{matrix}x=\dfrac{15+4\sqrt{3}}{10}\\x=\dfrac{15-4\sqrt{3}}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le1\\\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-4\le x< -2\)

\(\left(x-1\right)\left(1-9x^2\right)>=0\)

=>\(\left(x-1\right)\left(9x^2-1\right)< =0\)

TH1: \(\left\{{}\begin{matrix}x-1>=0\\9x^2-1< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=1\\x^2< =\dfrac{1}{9}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=1\\-\dfrac{1}{3}< =x< =\dfrac{1}{3}\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-1< =0\\9x^2-1>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =1\\x^2>=\dfrac{1}{9}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =1\\\left[{}\begin{matrix}x>=\dfrac{1}{3}\\x< =-\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3}< =x< =1\\x< =-\dfrac{1}{3}\end{matrix}\right.\)

a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1