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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
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
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
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 \(\left\{{}\begin{matrix}\sqrt{2x^2-2x+1}=a\\\sqrt{2x^2+2x+1}=b\end{matrix}\right.\)
\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\\3\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)
VH
23 tháng 9 2020
d, ĐKXĐ: \(x\ge-\frac{1}{4}\)
\(pt\Leftrightarrow4x^2+4x+2=2\sqrt{4x+1}\)
\(\Leftrightarrow4x^2+\left(4x+1-2\sqrt{4x+1}+1\right)=0\)
\(\Leftrightarrow4x^2+\left(\sqrt{4x+1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=0\\\sqrt{4x+1}-1=0\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)
VH
23 tháng 9 2020
a, ĐKXĐ: \(x\ge-1\)
\(pt\Leftrightarrow\sqrt{x+1}+\sqrt{x+8}=7\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x+8}\right)^2=49\)
\(\Leftrightarrow x+1+x+8+2\sqrt{\left(x+1\right)\left(x+8\right)}=49\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+8\right)}=20-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}20-x\ge0\\\left(x+1\right)\left(x+8\right)=\left(20-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le20\\49x=392\end{matrix}\right.\Leftrightarrow x=8\left(tm\right)\)
b, ĐKXĐ: \(x\ge-1\)
\(pt\Leftrightarrow\frac{x-3}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x-3}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
Do \(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}>0,\forall x\ge-1\)
Nên \(x=3\left(tm\right)\)
c, ĐKXĐ: \(x\ge-\frac{3}{2}\)
\(pt\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\)