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1.
ĐK: \(-1\le x\le4\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)
\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)
2.
ĐK:\(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)
\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)
\(PT\Leftrightarrow t=2x-12+t^2-2x\)
\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.
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 \(\sqrt{x^2+3x+4}-1\)
\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)
\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(-2\le x\le5\)
\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}-4=0\)
Đặt \(\sqrt{x+2}+\sqrt{5-x}=a>0\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}\)
\(\Rightarrow a+\frac{a^2-7}{2}-4=0\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}=1\)
\(\Leftrightarrow-x^2+3x+10=1\)
\(\Leftrightarrow x^2-3x-9=0\)
b/ \(\Leftrightarrow\sqrt{x+1}-\sqrt{4-x}+2\left(5+2\sqrt{\left(x+1\right)\left(4-x\right)}\right)=17\)
Đặt \(\sqrt{x+1}-\sqrt{4-x}=a\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{5-a^2}{2}\)
\(a+2\left(5+5-a^2\right)=17\)
\(\Leftrightarrow-2a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}-\sqrt{4-x}=-1\\\sqrt{x+1}-\sqrt{4-x}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+1=\sqrt{4-x}\\2\sqrt{x+1}=2\sqrt{4-x}+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2+2\sqrt{x+1}=4-x\\4x+4=25-4x+12\sqrt{4-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1-x\left(x\le1\right)\\12\sqrt{4-x}=8x-21\left(x\ge\frac{21}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left(1-x\right)^2\\144\left(4-x\right)=\left(8x-21\right)^2\end{matrix}\right.\)
c/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
\(a^2-1=3\left(a-1\right)\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-x^2}=\frac{a^2-1}{2}=0\\\sqrt{x-x^2}=\frac{a^2-1}{2}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\x-x^2=\frac{9}{4}\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{5+2x}=a\ge0\\\sqrt{5-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\left(3a-1\right)\left(3b-1\right)=16\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ab-\left(a+b\right)=5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3ab-5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\)
\(\Rightarrow\left(3ab-5\right)^2-2ab=10\)
\(\Leftrightarrow9\left(ab\right)^2-32ab+15=0\Rightarrow\left[{}\begin{matrix}ab=3\\ab=\frac{5}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(ab\right)^2=9\\\left(ab\right)^2=\frac{25}{81}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}25-4x^2=9\\25-4x^2=\frac{25}{81}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=\frac{500}{81}\end{matrix}\right.\)
e/ ĐKXĐ: \(-1\le x\le4\)
Tưởng nó giống câu c mà ko phải
\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(4-x\right)\left(x+1\right)}=5\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\) pt trở thành:
\(a+\frac{a^2-5}{2}=5\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)
\(\Leftrightarrow\sqrt{-x^2+3x+4}=2\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b/ĐKXĐ: \(0\le x\le4\)
\(\Leftrightarrow\left(3x-7\right)\sqrt{x\left(4-x\right)}+4-x=0\)
\(\Leftrightarrow\sqrt{4-x}\left[\left(3x-7\right)\sqrt{x}+\sqrt{4-x}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\\sqrt{4-x}=\left(7-3x\right)\sqrt{x}\left(x\le\frac{7}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow4-x=x\left(7-3x\right)^2\)
\(\Leftrightarrow4-x=x\left(9x^2-42x+49\right)\)
\(\Leftrightarrow9x^3-42x^2+50x-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(9x^2-24x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{4+\sqrt{14}}{3}>\frac{7}{3}\left(l\right)\\x=\frac{4-\sqrt{14}}{3}\end{matrix}\right.\)
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)
\(a,ĐK:1\le x\le3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)