Question

[Ly nguyễn gia]

(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)

Answer 1

[Nguyễn Việt Lâm]

6.

Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)

\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)

\(\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)

\(\Leftrightarrow5x^2+6x+5=16x^2\)

\(\Leftrightarrow11x^2-6x-5=0\)

\(\Rightarrow x=1\)

Answer 2

[Nguyễn Việt Lâm]

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 \(\sqrt{x^2+x+6}=t>0\)

\(t^2-\left(2x+1\right)t+6x-6=0\)

\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)