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5/
Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=a\ge0\\\sqrt{\frac{6}{x}-2x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=\frac{3}{x}\)
Pt trở thành:
\(a-1=\frac{a^2+b^2}{2}-b\)
\(\Leftrightarrow a^2+b^2-2a-2b+2=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=1\\\sqrt{\frac{6}{x}-2x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-x-3=0\\2x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{3}{2}\)
4/
ĐKXĐ: \(x\ge\frac{1}{5}\)
\(\Leftrightarrow\frac{4x-3}{\sqrt{5x-1}+\sqrt{x+2}}=\frac{4x-3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\frac{3}{4}\\\sqrt{5x-1}+\sqrt{x+2}=5\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x-1}-3+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}\right)=0\)
\(\Leftrightarrow x=2\)
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
f/
ĐKXĐ: ...
Đặt \(\sqrt{2-x}+\sqrt{x+2}=a>0\)
\(\Rightarrow a^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}\)
Phương trình trở thành:
\(a+\frac{a^2-4}{2}=2\)
\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}=0\)
\(\Rightarrow4-x^2=0\Rightarrow x=\pm2\)
e/ ĐKXĐ: ...
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:
\(a+\frac{a^2-5}{2}=5\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=2\)
\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Bạn ơi thứ nhất là làm ơi đặt câu hỏi hẳn hoi không thừa không thiếu đây bạn bài 1, 2 còn không cách ra đề bài thừa nhiều gây khó đọc và làm có khi là sai sẽ mất công người giải và chú ý là một câu hỏi thì chỉ nên hỏi một bài hoặc cụm câu liên quan tới nhau nha
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
ĐKXĐ: x\(\ge0\)
a/ \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-5\right)=x-17\)
\(\Leftrightarrow x-2\sqrt{x}-15=x-17\)
\(\Leftrightarrow-2\sqrt{x}=-2\Leftrightarrow x=1\)
b/ \(\frac{6-2\sqrt{x}+5}{6}=\frac{3-\sqrt{x}}{4}\)
\(\Leftrightarrow22-4\sqrt{x}=9-3\sqrt{x}\)
\(\Leftrightarrow x=169\)
c/ \(\Leftrightarrow x+6\sqrt{x}+9-x+3=0\)
\(\Leftrightarrow x=6\sqrt{x}=-12\) (vô lí)
Vây...