Giải phương trình $x^2-4x+6=\sqrt{2x^2-5x+3}+\sqrt{-3x^2+9x-5}$ - Phương trình - hệ phương trình - bất phương trình - Diễn đàn Toán học

bn giải cụ thể đc không ạ ?

c.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)

\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)

- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:

\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)

Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm

- Với \(x\le-5\) pt tương đương:

\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)

Do \(3-x>0\) pt trở thành:

\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)

\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)

\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))

\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)

\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)

a.

Kiểm tra lại đề, pt này không giải được

b.

ĐKXĐ: \(x\ge0\)

\(\sqrt{x\left(x+1\right)}-\sqrt{x}+1-\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)-\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x\ge\dfrac{1}{3}\)

\(\Leftrightarrow x^2+11x-3+2\sqrt{\left(x^2+2x\right)\left(9x-3\right)}=4x^2+13x+3\)

\(\Leftrightarrow2\sqrt{\left(x^2+2x\right)\left(9x-3\right)}=3x^2+2x+6\)

\(\Leftrightarrow2\sqrt{\left(3x+6\right)\left(3x^2-x\right)}=3x^2+2x+6\)

\(\Leftrightarrow\left(3x^2-x\right)-2\sqrt{\left(3x+6\right)\left(3x^2-x\right)}+3x+6=0\)

\(\Leftrightarrow\left(\sqrt{3x^2-x}-\sqrt{3x+6}\right)^2=0\)

\(\Leftrightarrow3x^2-x=3x+6\)

\(\Leftrightarrow3x^2-4x-6=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2+\sqrt{22}}{3}\\x=\dfrac{2-\sqrt{22}}{3}\left(loại\right)\end{matrix}\right.\)

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)

a/ ĐKXĐ: \(x^2+2x-6\ge0\)

\(\Leftrightarrow x^2+2x-6+\left(x-2\right)\sqrt{x^2+2x-6}=0\)

\(\Leftrightarrow\sqrt{x^2+2x-6}\left(\sqrt{x^2+2x-6}+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2+2x-6=\left(2-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)

Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(

c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\left(x^2+x\right)\left(x^2+x+2\right)}-\left(3-x\right)\sqrt{x^2+x}=0\)

\(\Leftrightarrow\sqrt{x^2+x}\left(\sqrt{x^2+x+2}-3+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\left(1\right)\\\sqrt{x^2+x+2}=3-x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)

d/

Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)

\(\Rightarrow\sqrt{x^2+3x+4}-1>0\)

Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)

\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)

\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)