Phương trình chứa căn

Lời giải:
ĐKXĐ: $x\leq 3$

PT $\Leftrightarrow \sqrt{15-x}=6-\sqrt{3-x}$

$\Rightarrow 15-x=36+3-x-12\sqrt{3-x}$ (bp hai vế)

$\Leftrightarrow 24=12\sqrt{3-x}$

$\Leftrightarrow 2=\sqrt{3-x}$

$\Rightarrow 3-x=4$

$\Leftrightarrow x=-1$ (thỏa mãn)

Thử lại thấy ok nên $x=-1$ là nghiệm duy nhất của pt.

b: ĐKXĐ: \(\left\{{}\begin{matrix}x< =4\\x\notin\left\{2;-\dfrac{1}{2}\right\}\end{matrix}\right.\)

Điều kiện: \(\left\{{}\begin{matrix}5x-10\ge0\\8-x\ge0\end{matrix}\right.\)\(\Leftrightarrow2\le x\le8\)

Bình phương hai vế ta được: \(5x-10=\left(8-x\right)^2\)

\(\Leftrightarrow5x-10=64-16x+x^2\)

\(\Leftrightarrow x^2-21x+74=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21+\sqrt{145}}{2}\left(loai\right)\\x=\dfrac{21-\sqrt{145}}{2}\left(nhan\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình là \(x=\dfrac{21-\sqrt{145}}{2}\)

3: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+2\left(x>=\dfrac{1}{2}\right)\\2x-1=-x-2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

\(1,ĐK:x\ge1\\ PT\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ 2,\Leftrightarrow2x-5=x^2-8x+16\left(x\ge4\right)\\ \Leftrightarrow x^2-10x+21=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\\ 3,\Leftrightarrow\left[{}\begin{matrix}2x-1=x+2\left(x\ge\dfrac{1}{2}\right)\\1-2x=x+2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)

\(ĐK:x\ge5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{x-5}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow4b^2-3a^2=x-20\)

\(PT\Leftrightarrow4b^2-3a^2+a+b+ab=0\\ \Leftrightarrow4ab+4b^2-3a^2-3ab+a+b=0\\ \Leftrightarrow4b\left(a+b\right)-3a\left(a+b\right)+\left(a+b\right)=0\\ \Leftrightarrow\left(a+b\right)\left(4b-3a+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=0\left(\text{loại do }a+b>0\right)\\4b-3a+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow4\sqrt{x-5}=3\sqrt{x}-1\\ \Leftrightarrow16x-80=9x-6\sqrt{x}+1\\ \Leftrightarrow7x+6\sqrt{x}-81=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-\dfrac{27}{7}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=9\left(nhận\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+7\ge0\\x-5=\left(3x+7\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{7}{3}\\9x^2+41x+54=0\left(vn\right)\end{matrix}\right.\)

Pt đã cho vô nghiệm

a: \(\Leftrightarrow x^2-12x+36-5x-6=0\)

\(\Leftrightarrow x-15=0\)

hay x=15