gpt : a) \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
b) \(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
c) \(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\frac{x+2}{x-2}}=-3\)
c) Đặt \(\left\{{}\begin{matrix}x-2=a\\x+2=b\end{matrix}\right.\)
\(pt\Leftrightarrow ab+4a\cdot\sqrt{\frac{b}{a}}=-3\)
\(\Leftrightarrow ab+\sqrt{\frac{16a^2\cdot b}{a}}+3=0\)
\(\Leftrightarrow ab+\sqrt{16ab}+3=0\)
\(\Leftrightarrow ab+4\sqrt{ab}+3=0\)
\(\Leftrightarrow ab+\sqrt{ab}+3\sqrt{ab}+3=0\)
\(\Leftrightarrow\sqrt{ab}\left(\sqrt{ab}+1\right)+3\left(\sqrt{ab}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}+1\right)=0\)
Dễ thấy \(VT>0\forall x\)
Do đó pt vô nghiệm
b)
Đặt \(\left\{\begin{matrix} \sqrt[3]{7-x}=a\\ \sqrt[3]{x-5}=b\end{matrix}\right.\). PT đã cho trở thành:
\(\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\)
\(\Leftrightarrow (a-b)\left(\frac{1}{a+b}-\frac{a^2+ab+b^2}{2}\right)=0\)
Nếu \(a-b=0\Leftrightarrow a=b\Leftrightarrow a^3=b^3\Leftrightarrow 7-x=x-5\)
\(\Leftrightarrow x=6\) (thỏa mãn)
Nếu \(\frac{1}{a+b}-\frac{a^2+ab+b^2}{2}=0\)
\(\Leftrightarrow (a^2+ab+b^2)(a+b)=2=a^3+b^3\)
\(\Leftrightarrow a^2b+ab^2=0\Leftrightarrow ab(a+b)=0\)
Hiển nhiên $a+b\neq 0$ (để biểu thức có nghĩa)
Do đó \(\left[\begin{matrix} a=0\\ b=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=7\\ x=5\end{matrix}\right.\)
Vậy........
Dạng căn thức thì nên thận trọng khi đặt nhân tử chung hoặc đưa 1 biểu thức ra ngoài căn hay đưa vào trong căn, vì nó liên quan đến dấu.
a/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x\le-3\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) \(\Rightarrow ptvn\)
- Với \(x\ge1\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\\sqrt{2x+6}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3x+5+2\sqrt{\left(2x+6\right)\left(x-1\right)}=4x+4\)
\(\Leftrightarrow2\sqrt{\left(2x+6\right)\left(x-1\right)}=x-1\)
\(\Leftrightarrow4\left(2x+6\right)\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+24=x-1\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-2\end{matrix}\right.\)
- Với \(x>2\Rightarrow\left\{{}\begin{matrix}VT>0\\VP< 0\end{matrix}\right.\) \(\Rightarrow\) vô nghiệm
- Với \(x\le-2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-4\left|x-2\right|\sqrt{\frac{x+2}{x-2}}+3=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-4\sqrt{\frac{\left(x-2\right)^2\left(x+2\right)}{x-2}}+3=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-4\sqrt{\left(x-2\right)\left(x+2\right)}+3=0\)
Đặt \(\sqrt{\left(x-2\right)\left(x+2\right)}=\sqrt{x^2-4}=a\ge0\)
\(a^2-4a+3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=1\\x^2-4=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=-\sqrt{13}\end{matrix}\right.\)
a) \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x^2+4x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\x-1=b\end{matrix}\right.\)
\(pt\Leftrightarrow\sqrt{2a\left(a+2\right)}+\sqrt{ab}-2a=0\)
\(\Leftrightarrow\sqrt{a}\left(\sqrt{2\left(a+2\right)}+\sqrt{b}-2\sqrt{a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\Leftrightarrow a=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\\\sqrt{2\left(a+2\right)}+\sqrt{b}-2\sqrt{a}=0\left(.\right)\end{matrix}\right.\)
\(\left(.\right)\Leftrightarrow\sqrt{2\left(a+2\right)}+\sqrt{b}=2\sqrt{a}\)
\(\Leftrightarrow2\left(a+2\right)+b=4a\)
\(\Leftrightarrow2a=b+4\)
\(\Leftrightarrow2\left(x+1\right)=x-1+4\)
\(\Leftrightarrow x=1\)
Vậy \(x\in\pm1\)
