NH
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K
Khách
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NH
14 tháng 4 2023
\(\dfrac{3}{1-x}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\left(x\ne1;x\ne-2\right)\)
\(< =>\dfrac{-3}{x-1}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)
\(< =>\dfrac{-3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{2\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)
suy ra
`-3(x+2)-2(x-1)=x+8`
`<=>-3x-6-2x+2=x+8`
`<=>-3x-2x-x=8+6-2`
`<=>-6x=12`
`<=>x=-2(ktmđk)`
Vậy phương trình vô nghiệm
14 tháng 4 2023
=>-3(x+2)-2x+2=x+8
=>-3x-6-2x+2=x+8
=>-5x-4=x+8
=>-6x=12
=>x=-2(loại)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2021
Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?
21 tháng 2 2021
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)
Suy ra: \(3x^2+7x-10=0\)
\(\Leftrightarrow3x^2-3x+10x-10=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)
TM
21 tháng 2 2021
a/ \(\dfrac{3x^2+7x-10}{x}=0\)
\(< =>3x^2+7x-10=0\)
\(< =>3x^2+10x-3x-10=0\)
\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)
\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)
\(< =>\left(3x+10\right)\left(x-1\right)=0\)
\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)
Vậy tập nghiệm của .....
PA
3
AM
7 tháng 2 2022
ĐKXĐ: \(x\ne1,-1\)
Ta có: \(\dfrac{x-2}{x+1}\ge\dfrac{3x+2}{x-1}-2\)
\(\dfrac{x-2}{x+1}\ge\dfrac{3x+2-2\left(x-1\right)}{x-1}\)
\(\dfrac{x-2}{x+1}-\dfrac{3x+2-2x+2}{x-1}\ge0\)
\(\dfrac{x-2}{x+1}-\dfrac{x+4}{x-1}\ge0\)
\(\dfrac{\left(x-2\right)\left(x-1\right)-\left(x-4\right)\left(x+1\right)}{x^2-1}\ge0\)
\(\dfrac{x^2-3x+2-x^2+3x+4}{x^2-1}\ge0\)
\(\dfrac{6}{x^2-1}\ge0\)
\(\Rightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\)(TM)
7 tháng 2 2022
\(BPT\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\ge\dfrac{\left(3x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x^2-x-2x+2-3x^2-3x-2x-2-2x^2-2\ge0\)
\(\Leftrightarrow-4x^2-8x-2\ge0\)
\(\Leftrightarrow x^2+2x+\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{1}{2}\ge0\)
Vậy bất phương trình luôn đúng \(\forall x\).
H
10 tháng 4 2021
\(\dfrac{15x-2}{4}-\dfrac{x^2+1}{3}>\dfrac{x\left(1-2x\right)}{6}+\dfrac{x-3}{2}\\ \Leftrightarrow3\left(15x-2\right)-4\left(x^2+1\right)>2x\left(1-2x\right)+6\left(x-3\right)\\ \Leftrightarrow45x-6-4x^2-4>2x-4x^2+6x-18\\ \Leftrightarrow45x-6x-2x>6+4-18\\ \Leftrightarrow37x>-8\\ \Leftrightarrow x>-\dfrac{8}{37}\)
MD
1
18 tháng 7 2021
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: \(\left\{{}\begin{matrix}-x< 0\\\left(x+2\right)\left(x+1\right)>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x< -2\\x>-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x>0\)
DH
1
HN
5 tháng 4 2022
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).
KV
5 tháng 4 2022
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 \(S=\varnothing\)
10 tháng 5 2023
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
ĐKXĐ: \(x\notin\left\{0;2\right\}\)
Ta có: \(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\)
\(\Leftrightarrow\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}>0\)
\(\Leftrightarrow\dfrac{x^2+x^2-4-2x^2+4x}{x\left(x-2\right)}>0\)
\(\Leftrightarrow\dfrac{4x-4}{x\left(x-2\right)}>0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}4x-4>0\\x\left(x-2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x>4\\\left[{}\begin{matrix}x>2\\x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\\left[{}\begin{matrix}x>2\\x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x>2\)
Kết hợp ĐKXĐ, ta được: x>2
Trường hợp 2:
\(\left\{{}\begin{matrix}4x-4< 0\\x\left(x-2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x< 4\\0< x< 2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 1\\0< x< 2\end{matrix}\right.\Leftrightarrow0< x< 1\)
Kết hợp ĐKXĐ, ta được: 0<x<1
Vậy: S={x|\(\left[{}\begin{matrix}x>2\\0< x< 1\end{matrix}\right.\)}