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:

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

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

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

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

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

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

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

\(\Leftrightarrow\sqrt{2x^2+16x+18}-\left(2x+4\right)+\sqrt{x^2-1}=0\)

\(\Leftrightarrow\dfrac{2x^2+16x+18-\left(4x^2+16x+16\right)}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)

\(\Leftrightarrow\dfrac{2x^2+16x+18-4x^2-16x-16}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)

\(\Leftrightarrow\dfrac{-2x^2+2}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)

\(\Leftrightarrow\dfrac{-2\left(x^2-1\right)}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(1-\dfrac{2\sqrt{x^2-1}}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}\right)=0\)

Tới đây đơn giản rồi

Sai đề k bạn ??

@Võ Hồng Phúc( phúc xịp)

\(ĐKXĐ:2x^2+16x+18\ge0;x^2-1\ge0\)

\(pt\Leftrightarrow\sqrt{x^2-1}=2x+4-\sqrt{2x^2+16x+18}\)(1)

\(\Leftrightarrow\sqrt{x^2-1}\left(\frac{2\sqrt{x^2-1}}{2x+4+\sqrt{2x^2+16x+18}}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-1}=0\\2\sqrt{x^2-1}=2x+4+\sqrt{2x^2+16x+18}\left(2\right)\end{cases}}\)

Lấy(1) + (2), ta được: \(3\sqrt{x^2-1}=4x+8\Leftrightarrow x=\frac{3\sqrt{57}-32}{7}\)

\(pt\Rightarrow\sqrt{x^2-1}=2x+4-\sqrt{2x^2+16x+18}\)

\(\Rightarrow\sqrt{\frac{1}{2}.\left(2x+4\right)^2-\frac{1}{2}.\left(2x^2+16x+18\right)}=2x+4-\sqrt{2x^2+16x+18}\)

Chia 2 vế cho \(\sqrt{2x^2+16x+18}\)

\(\Rightarrow\sqrt{\frac{\left(2x+4\right)^2}{2.\left(2x^2+16x+18\right)}-\frac{1}{2}}=\frac{2x+4}{\sqrt{2x^2+16x+18}}-1\)

Đặt \(\frac{2x+4}{\sqrt{2x^2+16x+18}}=a\)

\(\Rightarrow\sqrt{\frac{1}{2}a^2-\frac{1}{2}}=a-1\left(a\ge1\right)\)

Kết quả x = 1 nha , chính xác r nek

Đợi tẹo coi mình làm được không.

mk nghĩ bài này =1 

đặt \(x^2+2x=a\) , thay vào pt ta được:

\(\sqrt{3a+16}+\sqrt{a}=2\sqrt{a+4}\)

\(\Leftrightarrow\left(\sqrt{3a+16}\right)^2=\left(2\sqrt{a+4}-\sqrt{a}\right)^2\)

\(\Leftrightarrow3a+16=4a+16-4\sqrt{a\left(a+4\right)}+a\)

\(\Leftrightarrow\left(4\sqrt{a^2+4a}\right)^2=\left(2a\right)^2\)

\(\Leftrightarrow16a^2+64a=4a^2\)

\(\Leftrightarrow12a^2+64a=0\Leftrightarrow\orbr{\begin{cases}a=0\\a=-\frac{16}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x=0\\x^2+2x=-\frac{16}{3}\end{cases}}\)

Tự giải tiếp nhá

bạn đặt điều kiện cho a là \(a\ge-4\) rồi loại trường hợp \(a=\frac{-16}{3}\)

Mơn b nhé,Min......

\(\sqrt{2x+1}-\sqrt{18x+9}=\sqrt{32x+16}-18\left(đk:x\ge-\dfrac{1}{2}\right)\)

\(\Leftrightarrow\sqrt{2x+1}-3\sqrt{2x+1}-4\sqrt{2x+1}=-18\)

\(\Leftrightarrow6\sqrt{2x+1}=18\)

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

\(\Leftrightarrow2x+1=9\)

\(\Leftrightarrow x=4\left(tm\right)\)

\(\sqrt{2x+1}-9\sqrt{2x+1}-16\sqrt{2x+1}=-18\)

\(-24\sqrt{2x+1}=-18\)

\(\sqrt{2x+1}=\dfrac{3}{4}\)

\(\sqrt{\left(2x+1\right)^2}=\dfrac{9}{16}\)

\(2x+1=\dfrac{9}{16}\)

\(x=\dfrac{-7}{32}\)

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

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

\(\Leftrightarrow2x+2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8\)

\(\Leftrightarrow2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8-2x\)

\(\Leftrightarrow4\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]=64-32x+4x^2\)

\(\Leftrightarrow4x^2-32x+64=64-32x+4x^2+\)

\(\Leftrightarrow64=64\) (Đúng)

⇒ Phương trình có vô số nghiệm.

Vậy \(S=\mathbb R\).

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

ĐK: \(x\ge2\), PT tương đương với:

\(x+2\sqrt{2x-4}+2\sqrt{\left(x+2\sqrt{2x-4}\right)\left(x-2\sqrt{2x-4}\right)}+x-2\sqrt{2x-4}=8\)

\(\Leftrightarrow2x+2\sqrt{x^2-4\left(2x-4\right)}=8\)

\(\Leftrightarrow2x+2\sqrt{x^2-8x+16}=8\\ \Leftrightarrow x+\left|x-4\right|=8\)

Với x < 4 => \(x+4-x=8\)

\(\Leftrightarrow4=8\) (loại)

Với \(x\ge4\) => \(x+x-4=8\)

\(\Leftrightarrow x=6\) (thỏa mãn)