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a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a/
\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)
\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)
b/
\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)
\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)
c/
\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)
\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)
d/ \(x\ge\frac{3}{2}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)
e/
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)
f/
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
g/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x^2-x-2\\x^2-2x=-2x^2+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
e: =>-3<5x-12<3
=>9<5x<15
=>9/5<x<3
f: =>3x+15>=3 hoặc 3x+15<=-3
=>3x>=-12 hoặc 3x<=-18
=>x<=-6 hoặc x>=-4
b: =>(2x-7)(x-5)<=0
=>7/2<=x<=5
a: =>|x+3|=|2x-1|
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+3\\2x-1=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-2\end{matrix}\right.\Leftrightarrow x\in\left\{4;-\dfrac{2}{3}\right\}\)
b: \(\left|x^2-2x\right|=\left|2x^2-x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x-2=x^2-2x\\2x^2-x-2=-x^2+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+1\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1;-1;\dfrac{2}{3}\right\}\)
c: \(\left|3x^2-2x\right|=\left|6-x^2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow2x^2-x-3=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\)
=>x=3/2 hoặc x=-1
d: \(\left|2x^2-3x-5\right|=\left|x^2-4x-5\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=x^2-4x-5\\2x^2-3x-5=4x+5-x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\\3x^2-7x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\3x^2-10x+3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\\left(3x-10\right)\left(x+1\right)=0\end{matrix}\right.\)
hay \(x\in\left\{\dfrac{10}{3};-1\right\}\)
e: |5x+1|=|2x-3|
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=2x-3\\5x+1=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\7x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)