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c)
\(\sqrt{\left(x-1\right)^2}=2\)
x-1=2
x=3
d) \(\Leftrightarrow2+3\sqrt{x}+x=x+5\)
\(\Leftrightarrow3\sqrt{x}=3\)
<=> x=1
a)
\(\Leftrightarrow\sqrt{\left(x+2\right)}.\sqrt{\left(x-2\right)}-\sqrt{x+2}=0\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+2}=0\\\sqrt{x-2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
b)
\(\Leftrightarrow\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(\Leftrightarrow2\sqrt{x-2}+\sqrt{2}-\sqrt{2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{2}\)
\(\Leftrightarrow x-2=2\)
\(\Leftrightarrow x=4\)
2 phần kia mình đăng sau (dài quá r)
\(\left(\sqrt{3x+4}-\sqrt{3x+2}\right)\left(\sqrt{9x^2+18x+8}+1\right)=2\)
\(\Leftrightarrow\left(\sqrt{3x+4}-\sqrt{3x+2}\right)\left(\sqrt{\left(3x+4\right)\left(3x+2\right)}+1\right)=2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x+4}=a\\\sqrt{3x+2}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\), ta có hpt:
\(\left\{{}\begin{matrix}a^2-b^2=2\left(1\right)\\\left(a-b\right)\left(ab+1\right)=2\end{matrix}\right.\)
\(\Leftrightarrow a^2-b^2=\left(a-b\right)\left(ab+1\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(ab+1\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-ab-1\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(1-a\right)=0\)
* Trường hợp 1: \(a-b=0\Leftrightarrow a=b\)
\(\Rightarrow\sqrt{3x+4}=\sqrt{3x+2}\)
\(\Leftrightarrow0x=\sqrt{2}-2\)
=> Pt vô no
* Trường hợp 2: \(b-1=0\Leftrightarrow b=1\)
\(\Rightarrow\sqrt{3x+2}=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\left(n\right)\)
* Trường hợp 3: \(a-1=0\Leftrightarrow a=1\)
\(\Rightarrow\sqrt{3x+4}=1\)
\(\Rightarrow x=-1\left(l\right)\)
Vậy x = \(-\dfrac{1}{3}\)
