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Giải PT:
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 10 2019
a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)
b/
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)
\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)
\(\Leftrightarrow\left|x+2\right|-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
c/
\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)
\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2019
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)
Đặt \(\frac{\left|x-2\right|}{x-1}=a\)
\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
e/ ĐKXĐ: ...
Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2020
1.
\(\frac{x^2+2x+5}{x+4}-\left(x-3\right)\ge0\)
\(\Leftrightarrow\frac{x^2+2x+5-\left(x-3\right)\left(x+4\right)}{x+4}\ge0\)
\(\Leftrightarrow\frac{x+17}{x+4}\ge0\Rightarrow\left[{}\begin{matrix}x>-4\\x\le-12\end{matrix}\right.\)
2.
\(\frac{x^2-3x-1}{2-x}+x>0\)
\(\Leftrightarrow\frac{x^2-3x-1+x\left(2-x\right)}{2-x}>0\)
\(\Leftrightarrow\frac{-x-1}{2-x}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)
3.
\(\frac{3x-47}{3x-1}-\frac{4x-47}{2x-1}>0\)
\(\Leftrightarrow\frac{\left(3x-47\right)\left(2x-1\right)-\left(4x-47\right)\left(3x-1\right)}{\left(3x-1\right)\left(2x-1\right)}>0\)
\(\Leftrightarrow\frac{-6x\left(x-8\right)}{\left(3x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{1}{3}\\\frac{1}{2}< x< 8\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2020
4.
\(\frac{x\left(x+2\right)+9}{x+2}-4\ge0\)
\(\Leftrightarrow\frac{x^2+2x+9-4\left(x+2\right)}{x+2}\ge0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x+2}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x+2}\ge0\Rightarrow x>-2\)
5.
\(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Rightarrow\left[{}\begin{matrix}x\le-6\\1\le x< 2\\2< x< 7\\x=-2\end{matrix}\right.\)
6. Xem lại đề
HP
7 tháng 10 2020
c, ĐKXĐ: \(x\ne1\)
\(pt\Leftrightarrow\left(x-1\right)^2+\frac{9}{\left(x-1\right)^2}-7\left(x-1\right)+\frac{21}{x-1}=0\)
\(\Leftrightarrow\left(x-1-\frac{3}{x-1}\right)^2-7\left(x-1-\frac{3}{x-1}\right)+6=0\)
\(\Leftrightarrow\left(x+\frac{3}{x-1}-7\right)\left(x+\frac{3}{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{x-1}=7\\x+\frac{3}{x-1}=2\end{matrix}\right.\)
Nếu \(x+\frac{3}{x-1}=7\)
\(\Leftrightarrow x^2-8x+10=0\)
\(\Leftrightarrow x=4\pm\sqrt{6}\left(tm\right)\)
Nếu \(x+\frac{3}{x-1}=2\)
\(\Leftrightarrow x^2-3x+5=0\left(\text{vô nghiệm}\right)\)
Vậy pt fđã cho có hai nghiệm ...
ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4x+3>0\\log_{\frac{9}{16}}\left(x^2-4x+3\right)>0\end{matrix}\right.\)
\(\Rightarrow0< x^2-4x+3< 1\) \(\Rightarrow\left[{}\begin{matrix}2-\sqrt{2}< x< 1\\3< x< 2+\sqrt{2}\end{matrix}\right.\)
Khi đó:
\(\Leftrightarrow log_{\frac{9}{16}}\left(x^2-4x+3\right)\le1\)
\(\Leftrightarrow x^2-4x+3\ge\frac{9}{16}\)
\(\Leftrightarrow x^2-4x+\frac{39}{16}\ge0\Rightarrow\left[{}\begin{matrix}x\ge\frac{13}{4}\\x\le\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2-\sqrt{2}< x\le\frac{3}{4}\\\frac{13}{4}\le x< 2+\sqrt{2}\end{matrix}\right.\)