a) \(\sqrt{x+2}+\sqrt{x-1}=3x\) ( điều kiện: \(x\ge1\))
xét \(3x-\sqrt{x+2}-\sqrt{x-1}=A\)
\(2A=6x-2\sqrt{x+2}-2\sqrt{x-1}\)
\(2A=x+2-2\sqrt{x+2}+1+x-1-2\sqrt{x-1}+1+4x-3\)
\(2A=\left(\sqrt{x+2}-1\right)^2+\left(\sqrt{x-1}-1\right)^2+4x-3\)
ta có \(x\ge1\Leftrightarrow4x-3\ge1>0\)
\(\Rightarrow A>0\forall x\in R\)
vậy pt vô nghiệm
b) \(x^2+6=4\sqrt{\left(x+1\right)\left(x^2-3x+3\right)}\) (điều kiện \(x\ge-1\) )
\(\Leftrightarrow x^2-3x+3-4\sqrt{\left(x+1\right)\left(x^2-3x+3\right)}+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2-3x+3}-\sqrt{x+1}\right)\left(\sqrt{x^2-3x+3}-3\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+3}=\sqrt{x-1}\\\sqrt{x^2-3x+3}=3\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+3=x-1\\x^2-3x+3=9\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+4=0\\x^2-12x+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6+2\sqrt{6}\\x=6-2\sqrt{6}\end{matrix}\right.\)
a/ ĐKXĐ: \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{x-1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3x\\a^2-b^2=3\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)=\left(a+b\right)\left(a-b\right)x\Rightarrow a-b=\frac{1}{x}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3x\\a-b=\frac{1}{x}\end{matrix}\right.\) \(\Rightarrow2a=3x+\frac{1}{x}\)
\(\Rightarrow2\sqrt{x+2}=3x+\frac{1}{x}\Leftrightarrow x+2-2\sqrt{x+2}+1+2x+\frac{1}{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-1\right)^2+2x+\frac{1}{x}-3=0\)
Do \(2x+\frac{1}{x}-3=x+x+\frac{1}{x}-3\ge x+2\sqrt{x.\frac{1}{x}}-3\ge x-1\ge0\)
\(\Rightarrow VT\ge0\)
Dấu "=" xảy ra khi và chỉ khi
\(\left\{{}\begin{matrix}x=1\\\sqrt{x+2}-1=0\end{matrix}\right.\) \(\Rightarrow\) phương trình vô nghiệm
b/ĐKXĐ:...
\(x^2-3x+3+3\left(x+1\right)-4\sqrt{\left(x+1\right)\left(x^2-3x+3\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-3x+3}=b\end{matrix}\right.\)
\(\Rightarrow a^2+3b^2-4ab=0\Rightarrow\left(a-b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=3b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+3}=\sqrt{x+1}\\\sqrt{x^2-3x+3}=3\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+3=x+1\\x^2-3x+3=9\left(x+1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
