giai pt:
a) \(\frac{3x+\sqrt{x^2-x-1}}{x+1}=\frac{7}{3}\)
b) \(\frac{2}{2\sqrt{x^2-2x+1}}=\frac{1}{x-1}\)
c) \(\frac{6}{6-\sqrt{x}}+\frac{1}{\sqrt{x}}=1\)
d) \(\frac{2}{\sqrt{x-1}}+\sqrt{x-1}=\frac{3\sqrt{x-1}+1}{\sqrt{x-1}}-1\)
e) \(\sqrt{x+3-\sqrt{x-1}=2}\)
f) \(\sqrt{x^3+x^2+6x+28}=x+5\)
g) \(\sqrt{x^4-4x^3+14x-11}=1-x\)
ĐK: \(x^4-4x^3+14x-11\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)
e/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x+3-\sqrt{x-1}=4\)
\(\Leftrightarrow\sqrt{x-1}=x-1\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(\Leftrightarrow9x+3\sqrt{x^2-x-1}=7x+7\)
\(\Leftrightarrow3\sqrt{x^2-x-1}=7-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\9\left(x^2-x-1\right)=\left(7-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\5x^2+19x-58=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\frac{29}{5}\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne1\)
\(\Leftrightarrow\frac{1}{\sqrt{\left(x-1\right)^2}}=\frac{1}{x-1}\)
\(\Leftrightarrow\frac{1}{\left|x-1\right|}=\frac{1}{x-1}\)
\(\Rightarrow x-1>0\Rightarrow x>1\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne36\end{matrix}\right.\)
\(\Leftrightarrow6\sqrt{x}+6-\sqrt{x}=\left(6-\sqrt{x}\right)\sqrt{x}\)
\(\Leftrightarrow5\sqrt{x}+6=6\sqrt{x}-x\)
\(\Leftrightarrow x-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{23}{4}=0\)
Phương trình vô nghiệm
d/ ĐKXĐ: \(x>1\)
\(\Leftrightarrow2+x-1=3\sqrt{x-1}+1-\sqrt{x-1}\)
\(\Leftrightarrow2\sqrt{x-1}=x\)
\(\Leftrightarrow4\left(x-1\right)=x^2\)
\(\Leftrightarrow x^2-4x+4=0\Rightarrow x=2\)
Né câu b,c,d ra cho khỏi bị lặp :3
e, ĐK: \(x\ge1\)
\(PT\Leftrightarrow x+3-\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=\sqrt{x-1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-2x+1=x-1\end{matrix}\right.\Rightarrow\)\(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\left(tm\right)\)
d,\(ĐK:x\ge-2,673\)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3+x^2+6x+28=x^2+10x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left(x-1\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1+\sqrt{13}}{2}\\x=\frac{-1-\sqrt{13}}{2}\left(l\right)\end{matrix}\right.\)
g/
\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^4-4x^3+14x-11=\left(1-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=2\left(l\right)\\x=3\left(l\right)\end{matrix}\right.\)
