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a) \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
Vậy...
1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)
Vậy \(S=\left\{4;-1\right\}\)
\(3,x^4-5x^2+6=0\)
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành
\(t^2-5t+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)
\(\Rightarrow\)Pt ó 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)
\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)
Vậy \(S=\left\{\pm\sqrt{3}\right\}\)
T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(
\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)
\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)
\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)
\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)
\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
Đặt \(a=\sqrt[3]{3x+1};b=\sqrt[3]{3x-1}\)=>a3 -b3 =2
pt ,<=> a2 +b2 +ab =1 (1)
Mà a3 - b3 = ( a-b)(a2+b2+ab) =2
=> a -b =2 (*)
=> a2 +b2 -2ab =4 (2)
(1)(2)=> 3ab =-3 => ab =-1 => a(-b) =1 (**)
(*)(**)=> a ; -b là nghiệm của pt: x2 -2x+1=0 => a =-b =1
=> \(3x+1=1\Rightarrow x=0\)
và \(3x-1=-1\Rightarrow x=0\)
Vây x =0 (TM)