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\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{2x-9}=\sqrt[3]{x-5}+\sqrt[3]{4x-3}\)

Đặt \(\sqrt[3]{3x+1}=a;\sqrt[3]{2x-9}=b;\sqrt[3]{x-5}=c;\sqrt[3]{4x-3}=d\) ta được hệ:

\(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c+d\\\left(a+b\right)^3-3ab\left(a+b\right)=\left(c+d\right)^3-3cd\left(c+d\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+b=c+d=0\\\left[{}\begin{matrix}a+b=c+d\ne0\\ab=cd\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a^3+b^3=0\\a^3b^3=c^3d^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\\left(3x+1\right)\left(2x-9\right)=\left(4x-3\right)\left(x-5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\x^2-x-12=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x^3-1\ge0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)\ge0\) 

mà \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(2x^2+5x-1=7\sqrt{x^3-1}\Leftrightarrow2x^2+2x+2+3x-3=7\sqrt{x-1}\sqrt{x^2+x+1}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{x-1}\sqrt{x^2+x+1}\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{x^2+x+1}\end{matrix}\right.\left(a,b\ge0\right)\)

\(\Rightarrow\) pt trở thành \(2b^2+3a^2=7ab\Rightarrow2b^2-7ab+3a^2=0\)

\(\Rightarrow2b^2-6ab-ab+3a^2=0\Rightarrow2b\left(b-3a\right)-a\left(b-3a\right)=0\)

\(\Rightarrow\left(b-3a\right)\left(2b-a\right)=0\Rightarrow\left[{}\begin{matrix}b=3a\\2b=a\end{matrix}\right.\)

\(TH_1:b=3a\Rightarrow\sqrt{x^2+x+1}=3\sqrt{x-1}\)

\(\Rightarrow x^2+x+1=9\left(x-1\right)\Rightarrow x^2-8x+10=0\)

\(\Delta=\left(-8\right)^2-4.10=24\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{8-\sqrt{24}}{2}=4-\sqrt{6}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{8+\sqrt{24}}{2}=4+\sqrt{6}\end{matrix}\right.\)

\(TH_2:2b=a\Rightarrow2\sqrt{x^2+x+1}=\sqrt{x-1}\)

\(\Rightarrow4\left(x^2+x+1\right)=x-1\Rightarrow4x^2+3x+5=0\)

mà \(4x^2+3x+5=\left(2x\right)^2+2.2x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\dfrac{71}{16}=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}>0\)

\(\Rightarrow\) loại 

Vậy pt có tập nghiệm \(S=\left\{4+\sqrt{6};4-\sqrt{6}\right\}\)

Đk \(x\ge0\)

Pt \(\Leftrightarrow\dfrac{2x+1-3x}{\sqrt{2x+1}+\sqrt{3x}}=x-1\)

\(\Leftrightarrow\dfrac{1-x}{\sqrt{2x+1}+\sqrt{3x}}+\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}+1\right)=0\)

\(\Leftrightarrow1-x=0\)( vì \(\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}+1>0\) với mọi \(x\ge0\))

\(\Leftrightarrow x=1\)

Vậy S={1}

ĐK: 2x + 3 \(\ge\) 0; x+ 1 \(\ge\) 0  => x \(\ge\) -1

Đặt \(t=\sqrt{2x+3}+\sqrt{x+1}\left(t\ge0\right)\)

=> \(t^2=3x+4+2.\sqrt{\left(2x+3\right)\left(x+1\right)}=3x+4+2\sqrt{2x^2+5x+3}\)

PT đã cho trở thành: t = t ² - 20 <=> t² - t - 20 = 0 <=> t = 5 ; t = -4 

t = 5 thỏa mãn => \(\sqrt{2x+3}+\sqrt{x+1}=5\) (*)

Nhận xét : x = 3 là nghiệm của phương trình

+) x < 3 => \(\sqrt{2x+3}+\sqrt{x+1}\sqrt{9}+\sqrt{4}=5\)=>  x> 3 không là nghiệm của (*)

vậy PT có 1 nghiệm duy nhất x = 3

\(ĐKXĐ:x\ge-1\)

Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\\\sqrt{x+1}=b\end{cases}\left(a,b\ge0\right)\Rightarrow}a^2+b^2-4=3x\)

Phương trình đã cho trở thành :

\(a+b=a^2+b^2-4+2ab-16\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-20=0\)

\(\Leftrightarrow\left(a+b-5\right)\left(a+b+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b=5\\a+b=-4\end{cases}}\) \(\Leftrightarrow a+b=5\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=21-3x\)

\(\Leftrightarrow\hept{\begin{cases}21-3x\ge0\\4.\left(2x+3\right)\left(x+1\right)=\left(21-3x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le7\\4.\left(2x^2+5x+3\right)=441-126x+9x^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le7\\x^2-146x+429=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le7\\\left(x-3\right)\left(x-143\right)=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le7\\\orbr{\begin{cases}x=3\\x=143\end{cases}}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\le7\\\left(x-3\right)\left(x-143\right)=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\le7\\\orbr{\begin{cases}x=3\\x=143\end{cases}}\end{cases}}\)\(\Leftrightarrow x=3\) ( Thỏa mãn ĐKXĐ )

Vậy pt có nghiệm duy nhất \(x=3\)

Đặt \(\sqrt{x^2+1}=t>0\)

\(\Rightarrow\left(4x-1\right)t=2t^2-2x\)

\(\Leftrightarrow2t^2-\left(4x-1\right)t-2x=0\)

\(\Delta=\left(4x-1\right)^2+16x=\left(4x+1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{4x-1-\left(4x+1\right)}{4}=-\dfrac{1}{2}\left(loại\right)\\t=\dfrac{4x-1+4x+1}{4}=2x\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+1}=2x\) (\(x\ge0\))

\(\Leftrightarrow x^2+1=4x^2\)

\(\Rightarrow x=\dfrac{\sqrt{3}}{3}\)