a) \(x^2-5+\sqrt{x+5}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\sqrt{x+5}=0\)(tự làm tiếp)
b) Đề hơi sai sai
c) Mik chưa nghĩ ra
d) \(\left(\sqrt{1-2x}-1\right)+\left(\sqrt{1+2x}-1\right)+x^2=0\)
\(\frac{-2x}{\sqrt{1-2x}+1}+\frac{2x}{\sqrt{1+2x}+1}+x^2=0\)(tự lm tiếp)

\(\hept{\begin{cases}\sqrt{x^2+1}=a\\2x^2+2x+1=b\end{cases}}\)

\(\Rightarrow2\left(b-2a^2\right)+1=4x-1\)

\(\Rightarrow\left(2b-4a^2+1\right)a=b\)

\(\Leftrightarrow\left(2a-1\right)\left(b-2a^2-a\right)=0\)

Làm nôt

ĐKXĐ: x\(\ge0\)

Ta có: 

\(\left(x+3\sqrt{x}\right)\left(x+9\sqrt{x}+18\right)=168\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x+3}\right)\left(\sqrt{x}+6\right)=168\)

\(\Leftrightarrow\left(x+6\sqrt{x}\right)\left(x+6\sqrt{x}+9\right)=168\)

Đặt \(x+6\sqrt{x}=a\)\(\left(a\ge o\right)\). Khi đó:

\(a\left(a+9\right)=168\Leftrightarrow a^2+9a-168=0\)

Bn tu giải tiếp nhé

Xửa đề:

\(\left(x+1\right)\left(x+4\right)+3\left(x+4\right)\sqrt{\frac{x+1}{x+4}}-18=0\)

Xet \(x+4>0\)

\(\Rightarrow\left(x+1\right)\left(x+4\right)+3\sqrt{\left(x+1\right)\left(x+4\right)}-18=0\)

Đặt \(\sqrt{\left(x+1\right)\left(x+3\right)}=a\)

\(\Rightarrow a^2+3a-18=0\)

Trường hợp \(x+4< 0\)

Làm tương tự

\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)

\(\Rightarrow a^2-b^2=1\)

\(\Rightarrow a^2=1+b^2\)

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

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

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

\(\Leftrightarrow\left(x-\frac{1}{x}-1\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{x}-1=0\)

Làm nôt

ĐKXĐ : \(\hept{\begin{cases}x+3\ge0\\x+5\ge0\end{cases}\Leftrightarrow x\ge-3}\)

Đặt \(\hept{\begin{cases}\sqrt{x+3}=a\\\sqrt{x+5}=b\end{cases}\left(a;b\ge0\right)}\)

\(\Rightarrow\sqrt{x^2+8x+15}=ab\)

Pt có dạng ::

ab = 3a + 2b - 6

<=> 3a - ab + 2b - 6 = 0

<=> a( 3 - b ) + 2 ( b - 3 ) = 0

<=> ( a - 2 ) ( 3 - b ) = 0

Đến đây tự giải tiếp nha ! Muộn rồi ngủ ngon ... BYe !