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a,\(\left(3x-2\right)\left(x+3\right)=9x^2-4\\ \Leftrightarrow\left(3x-2\right)\left(x+3\right)-\left(3x-2\right)\left(3x+2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x+3-3x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(-2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b, ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{x-4}{x+2}-\dfrac{x+1}{x-2}=\dfrac{24}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{24}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2-6x+8-x^2-3x-2-24}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow-9x-18=0\\ \Leftrightarrow x=-2\left(ktm\right)\)
\(\dfrac{2x+2}{5}+\dfrac{3}{10}\)<\(\dfrac{3x-2}{4}\)
⇒ \(\dfrac{4\left(2x+2\right)}{20}+\dfrac{6}{20}-\dfrac{5\left(3x-2\right)}{20}\)<0
⇔\(8x+8+6-15x+10< 0\) ( vì 20>0)
⇔ \(-7x< -24\)
⇔\(x\)>\(\dfrac{24}{7}\)
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)
\(\dfrac{x-2}{2}+1\le\dfrac{x-1}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{6}+\dfrac{1.6}{6}\le\dfrac{2\left(x-1\right)}{6}\)
`<=> 3x - 6 + 6 <= 2x-2`
`<=> 3x <= 2x-2`
`<=> 3x -2x <= -2`
`<=> x <= -2`
\(\dfrac{x-2}{2}\)+1≤\(\dfrac{x-1}{3}\)
<=>\(\dfrac{3x-6}{6}\)+\(\dfrac{6}{6}\)≤\(\dfrac{2x-1}{6}\)
<=>3x-6+6≤2x-1
<=>x<-1
a: =>x^2-8x+16<x^2-8x
=>16<0(loại)
b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)
=>x+1/2>=5/3x-1
=>-2/3x>=-3/2
=>x<=3/2:2/3=9/4
c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)
=>21-3x<=8x-16
=>-11x<=-37
=>x>=37/11
Ta có: \(\dfrac{x}{x+2}< \dfrac{x}{x+1}\)
\(\Leftrightarrow\dfrac{x}{x+2}-\dfrac{x}{x+1}< 0\)
\(\Leftrightarrow\dfrac{x^2+x-x^2-2x}{\left(x+2\right)\left(x+1\right)}< 0\)
\(\Leftrightarrow\dfrac{-x}{\left(x+2\right)\cdot\left(x+1\right)}< 0\)
Trường hợp 1: \(\left\{{}\begin{matrix}-x>0\\\left(x+2\right)\left(x+1\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\-2< x< -1\end{matrix}\right.\Leftrightarrow-2< x< -1\)
Trường hợp 2: \(\left\{{}\begin{matrix}-x< 0\\\left(x+2\right)\left(x+1\right)>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x< -2\\x>-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x>0\)
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
=>(x+2-x-3)/(x+3)<0
=>-1/x+3<0
=>x+3>0
=>x>-3
ko hiểu là sao