a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)

\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)

Pt trở thành:

\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)

\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)

b.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)

Pt trở thành:

\(t=t^2-4-16\Leftrightarrow...\)

\(ĐKXĐ:x\ge-1\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}\) ta được 

\(a^2-a-20=0\Rightarrow\orbr{\begin{cases}a=5\\a=-4\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}-2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{2}{\sqrt{2x+3}+3}+\frac{1}{\sqrt{x+1}+2}\right)=0\)

\(\Rightarrow x=3\)

b ) \(ĐKXĐ:x\ge0\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x}+2x+1+2\sqrt{x^2+x}-2=0\)

Đặt \(\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}\)

\(\Rightarrow a+a^2-2=0\Rightarrow\orbr{\begin{cases}a=1\\a=-2\left(l\right)\end{cases}}\)

\(\Rightarrow\sqrt{x+1}+\sqrt{x}=1\)

Mà \(x\ge0\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\sqrt{x+1}\ge1\end{cases}\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1}\)

Dấu " = " xảy ra khi và chỉ khi \(x=0\)

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

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

Đặt \(t=\sqrt{2x+3}+\sqrt{x+1}\left(t\ge0\right)\)

\(\Rightarrow t^2=3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}\)

\(\Rightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=t^2-3x-4\)

Pt <=> \(t=3x+t^2-3x-4-16\)

\(\Leftrightarrow t^2-t-20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=21-3x\)

\(\Leftrightarrow x^2-146x+429=0\)

...

Câu b giải tương tự

\(ĐKXĐ:...\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2-4=3x+2\sqrt{2x^2+5x+3}\left(1\right)\)

Phương trình trở thành :

\(a=a^2-4-16\Leftrightarrow a^2-a-20=0\Rightarrow\orbr{\begin{cases}a=5\\a=-4\left(l\right)\end{cases}}\)

Thay vào (1)

\(\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{2}{\sqrt{2x+3}+3}+\frac{1}{\sqrt{x+1}+2}\right)=0\)

\(\Rightarrow x=3\)