Đk:\(x\ge1\)

\(pt\Leftrightarrow3\left(x-2\right)\sqrt{x-1}\sqrt{x^2+x+1}+18\left(x-1\right)=x\left(x^2+x+1\right)\)

Chia 2 vế của pt cho \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)ta đc:

\(3\left(x-2\right)\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}+\frac{18\left(x-1\right)}{x^2+x+1}=x\)

Đặt \(y=\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}\left(y\ge0\right)\) pt trở thành

\(3\left(x-2\right)y+18y^2-x=0\)

\(\Leftrightarrow\left(3y-1\right)\left(6y+x\right)=0\)

\(\Leftrightarrow3y-1=0\left(y\ge0;x\ge1\Rightarrow6y+x\ge1\right)\)

\(\Leftrightarrow y=\frac{1}{3}\)\(\Leftrightarrow\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}=\frac{1}{3}\)

\(\Leftrightarrow9\left(x-1\right)=x^2+x+1\)

\(\Leftrightarrow x^2-8x+10=0\)

\(\Leftrightarrow x=4\pm\sqrt{6}\)

Vậy...

\(\Leftrightarrow7\left(2x-1\right)-15x=-3x\)

=>14x-7-15x+3x=0

=>2x=7

hay x=7/2(nhận)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{7}{3x}-\dfrac{5}{2x-1}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}+\dfrac{5}{1-2x}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{1}{1-2x}-\dfrac{5}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{-4}{1-2x}\)

\(\Rightarrow-4.3x=7\left(1-2x\right)\)

\(\Rightarrow-12x=7-14x\)

\(\Rightarrow-12x+14x=7\)

\(\Rightarrow2x=7\)

\(\Rightarrow x=\dfrac{7}{2}\left(tm\right)\)

PT⇔7/3x=5/2x-1 - 1/2x-1

     <=>7/3x=4/2x-1

     <=>7(2x-1)=4.3x

      <=>14x-7=12x

       <=>14x-12x=7

       <=>2x=7

       <=>x=7/2

Vậy tập nghiệm của pt là S={7/2}

\(\sqrt{4x-8}-\sqrt{x-2}=2.\)

ĐK \(x\ge2\)

PT<=> \(2\sqrt{x-2}-\sqrt{x-2}=2\)

<=> \(\sqrt{x-2}=2\)

<=> x-2=4

<=> x=6 (t/m)

Vậ pt có nghiệm x=6

mơn bn nha

\(ĐK:x\ge5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{x-5}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow4b^2-3a^2=x-20\)

\(PT\Leftrightarrow4b^2-3a^2+a+b+ab=0\\ \Leftrightarrow4ab+4b^2-3a^2-3ab+a+b=0\\ \Leftrightarrow4b\left(a+b\right)-3a\left(a+b\right)+\left(a+b\right)=0\\ \Leftrightarrow\left(a+b\right)\left(4b-3a+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=0\left(\text{loại do }a+b>0\right)\\4b-3a+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow4\sqrt{x-5}=3\sqrt{x}-1\\ \Leftrightarrow16x-80=9x-6\sqrt{x}+1\\ \Leftrightarrow7x+6\sqrt{x}-81=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-\dfrac{27}{7}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=9\left(nhận\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^3-3\left|y+1\right|=18\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x^3=40\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3=8\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3=8\\\left|y+1\right|=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\end{matrix}\right.\)

Đáp án A