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a ) \(\left(2x-1\right)^2+\left(x+4\right)^2+5=2x\left(x+1\right)+\left(x+2\right)^2+2x^2-2x+18\)
\(\Leftrightarrow4x^2-4x+1+x^2+8x+16+5=2x^2+2x+x^2+4x+4+2x^2-2x+18\)
\(\Leftrightarrow5x^2+4x+22=5x^2+4x+22\)
=> PT có vô số nghiệm
b ) \(\dfrac{5x-7}{4}-\dfrac{9x-4}{5}=-x-\dfrac{19-9x}{20}\)
\(\Leftrightarrow\dfrac{25x-35-36x+16}{20}=\dfrac{-20x-19+9x}{20}\)
\(\Leftrightarrow\dfrac{-11x-19}{20}=\dfrac{-11x-19}{20}\)
=> PT có vô số nghiệm
c ) \(\left|y-3\right|=y-3\)
TH 1 : \(y\ge3\)
\(\Rightarrow y-3\ge0\Rightarrow\left|y-3\right|=y-3\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow y-3=y-3\)
Nên : \(y\ge3\) , PT vô số nghiệm
TH 2 : \(y< 3\Rightarrow y-3< 0\Rightarrow\left|y-3\right|=3-y\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow3-y=y-3\)
\(\Rightarrow3-y-y+3=0\)
\(\Rightarrow6-2y=0\)
\(\Rightarrow y=3\) ( L ; do y < 3 )
Vậy \(y\ge3\) thì PT vô số nghiệm
1: =>2x-5=4 hoặc 2x-5=-4
=>2x=9 hoặc 2x=1
=>x=9/2hoặc x=1/2
2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)
3: \(\Leftrightarrow\left|5x-3\right|=x+5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)
a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)
⇔\(x=\dfrac{21-9x+10x+4}{12}\)
⇔x=\(\dfrac{x+25}{12}\)
⇔12x=x+25
⇔x=\(\dfrac{25}{11}\)
Vậy pt đã cho có n0 là S=\(\left\{\dfrac{25}{11}\right\}\)
b.ĐKXĐ:x≠-2;x≠2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)
⇔x2-7x-2=2x-22
⇔x2-9x+20=0
⇔(x-4)(x-5)=0
⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy pt đã cho có n0 là S={4;5}
\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)
\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)
\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)
Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)
\(\dfrac{-5x+7-2x+3-3x+4}{2-x}=\dfrac{-10x+14}{2-x}=\dfrac{-2\left(5x-7\right)}{2-x}=\dfrac{2\left(5x-7\right)}{x-2}.\)