ĐKXĐ: ...

\(\sqrt[3]{x^3+5x^2}-x-2+x+1-\sqrt{\dfrac{5x^2-2}{6}}=0\)

\(\Leftrightarrow\dfrac{x^3+5x^2-\left(x+2\right)^3}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}+\dfrac{\left(x+1\right)^2-\dfrac{5x^2-2}{6}}{x+1+\sqrt{\dfrac{5x^2-2}{6}}}=0\)

\(\Leftrightarrow\left(x^2+12x+8\right)\left(\dfrac{1}{6\left(x+1\right)+\sqrt{6\left(5x^2-2\right)}}-\dfrac{1}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}\right)=0\)

\(\Leftrightarrow x^2+12x+8=0\)

Bạn tham khảo thêm ở link sau:

https://hoc24.vn/cau-hoi/giai-phuong-trinhsqrt3x2-5x1-sqrtx2-2sqrt3leftx2-x-1right-sqrtx2-3x4.167769342831

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

ĐK: \(x\ge1\)

Đặt \(\sqrt{3x-2}+2\sqrt{x-1}=t\left(t\ge1\right)\)

\(pt\Leftrightarrow3t=t^2-4\)

\(\Leftrightarrow t^2-3t-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-1\left(l\right)\end{matrix}\right.\)

\(t=4\Leftrightarrow\sqrt{3x-2}+2\sqrt{x-1}=4\)

\(\Leftrightarrow7x-6+4\sqrt{\left(3x-2\right)\left(x-1\right)}=16\)

\(\Leftrightarrow4\sqrt{3x^2-5x+2}=22-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}48x^2-80x+32=484+49x^2-308x\\22-7x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}452+x^2-228x=0\\x\le\dfrac{22}{7}\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\4-x>=0\end{matrix}\right.\)

=>2<=x<=4

\(\sqrt{x-2}-\sqrt{4-x}=2x^2-5x-3\)

=>\(\sqrt{x-2}-1+1-\sqrt{4-x}=2x^2-6x+x-3\)

=>\(\dfrac{x-2-1}{\sqrt{x-2}+1}+\dfrac{1-4+x}{1+\sqrt{4-x}}=\left(x-3\right)\left(2x+1\right)\)

=>\(\left(x-3\right)\left(\dfrac{1}{\sqrt{x-2}+1}+\dfrac{1}{1+\sqrt{4-x}}-2x-1\right)=0\)

=>x-3=0

=>x=3(nhận)

`x^2+\sqrt{2x+1}+sqrt{x-3}=5x`

Bài này dùng pp liên hợp với đk của x là `x>=3`

`pt<=>x^2-16+\sqrt{2x+1}-3+\sqrt{x-3}-1=5x-20`

`<=>(x-4)(x+4)+(2x-8)/(\sqrt{2x+1}+3)+(x-4)/(\sqrt{x-3}+1)=5(x-4)`

`<=>(x-4)(x+4+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)-5)=0`

`<=>(x-4)(x-1+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1))=0`

Vì `x>=3=>x-1>=2>0`

Mà `2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)>0`

`=>x-1+2/(\sqrt{2x+1}+3)+1/(\sqrt{x-3}+1)>0`

`=>x-4=0<=>x=4(tm)`

Vậy `S={4}`