a/ ĐKXĐ: ....
\(\Leftrightarrow2x^2+2x+4+2x-4=5\sqrt{\left(x-2\right)\left(x^2+x+2\right)}\)
\(\Leftrightarrow2\left(x^2+x+2\right)+2\left(x-2\right)=5\sqrt{\left(x-2\right)\left(x^2+x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a\\\sqrt{x-2}=b\end{matrix}\right.\)
\(\Leftrightarrow2a^2+2b^2=5ab\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\sqrt{x-2}\\2\sqrt{x^2+x+2}=\sqrt{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\left(x-2\right)\\4\left(x^2+x+2\right)=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+10=0\\4x^2+3x+10=0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: ....
\(\Leftrightarrow2x^2-x+1=\sqrt{4x^4+4x^2+1-4x^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2+1\right)^2-\left(2x\right)^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
\(\Leftrightarrow\frac{3}{4}\left(2x^2-2x+1\right)+\frac{1}{4}\left(2x^2+2x+1\right)=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2-2x+1}=a\\\sqrt{2x^2+2x+1}=b\end{matrix}\right.\)
\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\\3\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)
c/
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2\right)^2+16x^2+1-16x^2}\)
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2+1\right)^2-16x^2}\)
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)}\)
\(\Leftrightarrow3\left(8x^2+4x+1\right)-2\left(8x^2-4x+1\right)=\sqrt{\left(8x^2+4x+1\right)\left(8x^2-4x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{8x^2+4x+1}=a>0\\\sqrt{8x^2-4x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow3a^2-2b^2=ab\Leftrightarrow3a^2-ab-2b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a+2b\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow\sqrt{8x^2+4x+1}=\sqrt{8x^2-4x+1}\)
\(\Leftrightarrow8x^2+4x+1=8x^2-4x+1\)
\(\Leftrightarrow x=0\)
d/ Phân tích cái căn kia trước:
\(x^4-8x+63=x^4+16x^2+64-16x^2-8x-1\)
\(=\left(x^2+8\right)^2-\left(4x+1\right)^2=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
Vậy ta có:
\(3x^2-4x+23=3\sqrt{\left(x^2+4x+9\right)\left(x^2-4x+7\right)}\)
\(\Leftrightarrow x^2+4x+9+2\left(x^2-4x+7\right)=3\sqrt{\left(x^2+4x+9\right)\left(x^2-4x+7\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+4x+9}=a>0\\\sqrt{x^2-4x+7}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2+2b^2=3ab\Leftrightarrow a^2-3ab+2b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+4x+9}=\sqrt{x^2-4x+7}\\\sqrt{x^2+4x+9}=2\sqrt{x^2-4x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+9=x^2-4x+7\\x^2+4x+9=4\left(x^2-4x+7\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
e/
Bạn xem lại đề bài, trong căn kia là \(16x^4+4x+1\) hay \(16x^4+4x^2+1\)
Biểu thức thế này thì không phân tích được
