Giải bất pt sau
\(\left(x-1\right)\left(3x^2+9x-12\right)< 0\)
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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
\(\Leftrightarrow\left(x^2-3x+1\right)\left(x+1\right)\left(x+2\right)\left(x-4\right)\left(x-5\right)=-30\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^2-3x-4\right)\left(x^2-3x-5\right)=-30\)
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
sorry b, phải là cái này nha