Nếu bạn thiếu số 2 bên cạnh $\sqrt{2x^2+5x+3}$ thì có thể tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/tim-x-sao-cho-sqrt2x3sqrtx13x2sqrt2x25x3-16.235781793134

Bạn kiểm tra lại đề bài

Đặt \(\sqrt{x^2+1}=t>0\)

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

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

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

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

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

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

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

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}

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

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: \(x\ge1\)

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

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)-7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}=0\)

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

\(\Rightarrow2a^2+3b^2-7ab=0\)

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

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

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

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0\)

Đặt \(\sqrt{x-2}-\sqrt{x+2}=a< 0\)

\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\)

Phương trình trở thành:

\(a+a^2-2=0\Leftrightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=-2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}=-2\)

\(\Leftrightarrow\sqrt{x-2}+2=\sqrt{x+2}\)

\(\Leftrightarrow x+2+4\sqrt{x-2}=x+2\)

\(\Leftrightarrow\sqrt{x-2}=0\)

e sửa lại câu 1 ạ: 2\(\text{x^2+5x−1=7√x3−1}\)

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

a) ĐK : \(x\ge\frac{2}{3}\)\(\sqrt{3x-2}-\sqrt{x+7}=1\Leftrightarrow3x-2-2\sqrt{\left(3x-2\right)\left(x+7\right)}+x+7=1\)

\(\Leftrightarrow4x+5-1=2\sqrt{3x^2+19x-14}\Leftrightarrow2x+2=\sqrt{3x^2+19x-14}\)

\(\Leftrightarrow4x^2+8x+4=3x^2+19x-14\)

\(\Leftrightarrow x^2-11x+18=0\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

b) ĐK \(x\ge-\frac{1}{5}\)\(\sqrt{14x+7}-\sqrt{2x+3}=\sqrt{5x+1}\Leftrightarrow14x+7+2x+3-5x-1-2\sqrt{28x^2+42x+14x+21}=0\)

\(\Leftrightarrow11x+9=2\sqrt{28x^2+56x+21}\Leftrightarrow121x^2+81+198x=112x^2+224x+84\)

\(\Leftrightarrow9x^2-26x-3=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{1}{9}\left(loai\right)\end{matrix}\right.\)

c) \(\sqrt{x^2+2x+6}-\sqrt{x^2+x+2}=1\)

\(\Leftrightarrow x^2+2x+6=x^2+x+2+1+2\sqrt{x^2+x+2}\)

\(\Leftrightarrow x+3=2\sqrt{x^2+x+2}\)

\(\Leftrightarrow x^2+6x+9=4x^2+4x+8\)

\(\Leftrightarrow3x^2-2x-1=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\frac{1}{3}\left(tm\right)\end{matrix}\right.\)