a. ĐKXĐ: \(x\le\frac{-2-\sqrt{2}}{2};x\ge\frac{-2+\sqrt{2}}{2}\)

\(pt\Leftrightarrow2\sqrt{2x^2+4x+1}=2-2x^2-4x\)

\(\Leftrightarrow2x^2+4x+1+2\sqrt{2x^2+4x+1}+1=0\)

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

\(\Leftrightarrow\sqrt{2x^2+4x+1}+1=0\)

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

\(\Rightarrow\text{pt vô nghiệm}\)

b. ĐKXĐ: \(x\le-4;x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t>0\right)\)

\(\Leftrightarrow t^2=2x+2\sqrt{x^2-16}\)

pt đã cho tương đương:

\(t=t^2\)

\(\Leftrightarrow t=1\) \(\left(\text{Vì }t>0\right)\)

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

\(\Leftrightarrow2x+2\sqrt{x^2-16}=1\)

\(\Leftrightarrow2\sqrt{x^2-16}=1-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-16\right)=\left(1-2x\right)^2\\1-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{65}{4}\\x\le\frac{1}{2}\end{matrix}\right.\Rightarrow\text{vô nghiệm}\)

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

=>2x-1=0 hoặc 2x+1=4

=>2x=1 hoặc 2x=3

=>x=3/2 hoặc x=1/2

b: \(\Leftrightarrow3x+2=2\left(x+2\right)\)

=>3x+2=2x+4

=>x=2(nhận)

mọi người jup mình giải đi khó wá

1 bài thui cx đc

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ge0\)

\(\sqrt{2x^2+13x+5}-5\sqrt{x}+\sqrt{2x^2-3x+5}-3\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2-12x+5}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{2x^2-12x+5}{\sqrt{2x^2-3x+5}+3\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-12x+5\right)\left(\dfrac{1}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-3x+5}+3\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-12x+5=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x^2\ge\dfrac{4}{3}\)

\(\sqrt{x^2-\dfrac{4}{3}}+\sqrt{4x^2-4}-x=0\)

\(\Leftrightarrow\sqrt{\dfrac{3x^2-4}{3}}+\dfrac{3x^2-4}{\sqrt{4x^2-4}+x}=0\)

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

\(\Leftrightarrow3x^2-4=0\)

\(\Leftrightarrow...\)