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\(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=b\end{matrix}\right.\) \(\Rightarrow a^3-b^3=2\) (1)
\(a^2+b^2+ab=1\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab\right)=a-b\)
\(\Leftrightarrow a^3-b^3=a-b\) (2)
Từ (1) và (2) \(\Rightarrow a-b=2\Rightarrow b=a-2\)
\(\Rightarrow a^2+\left(a-2\right)^2+a\left(a-2\right)=1\Leftrightarrow3a^2-6a+3=0\Rightarrow a=1\)
\(\Rightarrow\sqrt[3]{3x+1}=1\Rightarrow3x+1=1\Rightarrow x=0\)
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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)
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ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\frac{9x^2\left(1+\sqrt{1+3x}\right)^2}{\left(1-\sqrt{1+3x}\right)^2\left(1+\sqrt{1+3x}\right)^2}=3x+1\)
\(\Leftrightarrow\frac{9x^2\left(1+\sqrt{1+3x}\right)^2}{9x^2}=3x+1\)
\(\Leftrightarrow\left(1+\sqrt{1+3x}\right)^2=3x+1\)
Đặt \(\sqrt{3x+1}=t\ge0\)
\(\left(t+1\right)^2=t^2\Leftrightarrow2t+1=0\Rightarrow t=-\frac{1}{2}< 0\left(l\right)\)
Vậy pt đã cho vô nghiệm
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a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
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c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
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Lời giải:
$G=1-\sqrt{(3x-1)^2}+(3x-1)^2=1-|3x-1|+|3x-1|^2$
Đặt $|3x-1|=a$ với $a\geq 0$
Ta cần tìm GTNN của $G=1-a+a^2$
Có: $G=(a-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$ với mọi $a\geq 0$
Do đó gtnn của $G$ là $\frac{3}{4}$
\(9x^2-\left(3x+2\right)\sqrt{3x-1}+2=3x\left(1\right)\)\(\left(x\ge\frac{1}{3}\right)\)
Đặt \(\sqrt{3x-1}=a\ge0\)
\(\Rightarrow\hept{\begin{cases}3x=a^2+1\\3x+2=a^2+3\\3x-1=a^2\end{cases}}\)
Pt (1) \(\Leftrightarrow3x\left(3x-1\right)-\left(3x+2\right)\sqrt{3x-1}+2=0\)
\(\Leftrightarrow\left(a^2+1\right)a^2-\left(a^2+3\right)a+2=0\)
\(\Leftrightarrow a^3+a^2-a^3-3a+2=0\)
\(\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}}\)
TH1: a=1
\(\Rightarrow\sqrt{3x-1}=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow x=\frac{2}{3}\left(tm\right)\)
TH2: a=2
\(\Rightarrow\sqrt{3x-1}=2\)
\(\Leftrightarrow3x-1=4\)
\(\Leftrightarrow x=\frac{5}{3}\left(tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{\frac{2}{3};\frac{5}{3}\right\}\)