(Gấp xin hãy giải hộ vs ạ chiều e học rùi . E xin cảm ơn) GIẢI PHƯƠNG TRÌNH SAU
1)\(\frac{3}{\sqrt{x-2}+3}\)-\(\frac{1}{\sqrt{x+6}+3}\)=2
2)\(\sqrt{x^2+2x}\)+\(\sqrt{2x-1}\)=\(\sqrt{3x^2+4x+1}\)
4) ( 3x+1).\(\sqrt{2x^2-1}\)=5x2+\(\frac{3x}{2}\)
5) x2+7x=(2x+1).\(\sqrt{x^2+x+6}\)
6) \(\sqrt{5x^2+6x+5}\). (5x2+6x++6)=4x. (16x2+1)
1. ĐKXĐ: ...
Do \(\sqrt{x-2}\ge0\Rightarrow\frac{3}{\sqrt{x-2}+3}\le1\)
\(\frac{1}{\sqrt{x+6}+3}>0\)
\(\Rightarrow VT< 1-0=1< 2\)
Pt vô nghiệm
2.
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)ta được:
\(a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow a^2+b^2+2ab=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\) (nghiệm xấu quá bạn coi lại đề)
\(\Leftrightarrow\left(a-\frac{\sqrt{5}+1}{2}b\right)\left(a+\frac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\frac{\sqrt{5}+1}{2}b\)
\(\Leftrightarrow x^2+2x=\frac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-2\left(\frac{1+\sqrt{5}}{2}\right)x+\frac{3+\sqrt{5}}{2}=0\)
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
