\(ĐK:\left\{{}\begin{matrix}x\le\dfrac{1}{2};4\le x\\\dfrac{1}{2}\le x\\x\le-11;\dfrac{1}{2}\le x\end{matrix}\right.\Leftrightarrow x\le-11;4\le x\)

\(PT\Leftrightarrow\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\\ \Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}-\sqrt{x+11}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{x-4}-\sqrt{x+11}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x-4+x+11-2\sqrt{x^2+7x-44}=9\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{x^2+7x-44}=2x-2\\ \Leftrightarrow\sqrt{x^2+7x-44}=x-1\\ \Leftrightarrow x^2+7x-44=x^2-2x+1\\ \Leftrightarrow9x=45\Leftrightarrow x=5\left(tm\right)\)

Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)

https://hoc24.vn/cau-hoi/giai-pt-sqrt2x2-9x43sqrt2x-1sqrt2x221x-11.2005877637936

làm r nha :vv

Từ pt suy ra \(x\ge0\).

PT \(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}=2x\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+\left|\sqrt{2x-1}+1\right|=2x\). (*)

+) \(\sqrt{2x-1}-1\ge0\Leftrightarrow x\ge1\): Khi đó (*) tương đương \(2\sqrt{2x-1}=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) (thoả mãn)

+) \(\sqrt{2x-1}-1< 0\Leftrightarrow x< 1\): Khi đó (*) tương đương \(2=2x\Leftrightarrow x=1\), vô lí.

Vậy x = 1

ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`

`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`

`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`

`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`

`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`

`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`

(1) `<=> x-1=2\sqrt((x+3)(x-1))`

`<=>x^2-2x+1=4(x+3)(x-1)`

`<=>x=1\ `(TM)

Vậy `S={\pm 1}`.

\(ĐK:x\le-3;x\ge-1\)

\(PT\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)

Vậy \(S=\left\{-1;1\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\-4+\sqrt{7}\le x\le-1\end{matrix}\right.\)

Khi x thỏa ĐKXĐ, vế phải luôn dương, bình phương 2 vế ta được:

\(\Leftrightarrow3x^2+16x+17+2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=4x^2+16x+16\)

\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=x^2-1\)

\(\Leftrightarrow4\left(x^2-1\right)\left(2x^2+16x+18\right)=\left(x^2-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\4\left(2x^2+16x+18\right)=x^2-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\7x^2+64x+73=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-32+3\sqrt{57}}{7}\\x=\dfrac{-32-3\sqrt{57}}{7}\left(loại\right)\end{matrix}\right.\)

Bạn tham khảo nhé

Giải Phương Trình: (x+1)\(\sqrt{2x^2-2x}\) = 2x2-3x-2 - Hoc24

TK: https://hoc24.vn/cau-hoi/giai-phuong-trinh-x1sqrt2x2-2x-2x2-3x-2.261337627197