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a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)
\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)
Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)
Phương trình trở thành:
\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)
\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)
Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(
b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)
Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)
Pt trở thành:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)
\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)
Đặt \(x\sqrt{2x^2+4}=a\) ta được:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
a)\(ĐK:-3\le x\le6\)
\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)
\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)
b/ ĐKXĐ: \(x\ge7\)
\(\sqrt{3x-2}=1+\sqrt{x-7}\)
\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)
\(\Leftrightarrow x+2=\sqrt{x-7}\)
\(\Leftrightarrow x^2+4x+4=x-7\)
\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)
c/ ĐKXĐ: \(x\ge1;x\ne50\)
\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)
\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)
\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))
\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
a/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+5x+2}=2\sqrt{2x^2+5x-6}\)
\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)
\(\Leftrightarrow6x^2+15x-26=0\)
b/ ĐKXĐ: ...
Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)
\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)
c/ĐKXĐ: ...
\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)
Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)
\(\Leftrightarrow6x^2-12x-42=0\)
d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)
Đặt \(\sqrt{x^2+x+4}=a>0\)
\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+4}=2\Rightarrow x^2+x=0\)
e/ \(\Leftrightarrow x^2+2x+\sqrt{3x^2+6x+4}-2=0\)
Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)
\(\frac{a^2-4}{3}+a-2=0\)
\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)
a, ĐK: \(6x^2-12x+7\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)
Vậy...