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1)
\(x-\dfrac{x-1}{3}+\dfrac{x+2}{6}>\dfrac{2x}{5}+5\\ \Leftrightarrow x-\dfrac{x-1}{3}+\dfrac{x+2}{6}-\dfrac{2x}{5}-5>0\\ \Leftrightarrow\dfrac{30x-10\left(x-1\right)+5\left(x+2\right)-2x\cdot6-5\cdot30}{30}>0\\ \Leftrightarrow30x-10x+10+5x+10-12x-150>0\\ \Leftrightarrow30x-10x=5x-12x>-10-10+150\\ \Leftrightarrow13x>130\\ \Leftrightarrow13x\cdot\dfrac{1}{13}>130\cdot\dfrac{1}{13}\\ \Leftrightarrow x>10\)
Vậy tập ngiệm của bât hương trình là {x/x>10}
mình mới học đến đây nên cách giải còn dài, thông cảm nha
2)
\(\dfrac{2x+6}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{x+3}{3}-\dfrac{x-2}{9}-1< 0\\ \Leftrightarrow\dfrac{3\left(x+3\right)-x+2-9}{9}< 0\\ \Leftrightarrow3x+9-x+2-9< 0\\ \Leftrightarrow3x-x< -9+9-2\\ \Leftrightarrow2x< -2\\ \Leftrightarrow2x\cdot\dfrac{1}{2}< -2\cdot\dfrac{1}{2}\Leftrightarrow x< -1\)
Vậy tập nghiệm của bất phương trình là {x/x<-1}
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
ĐK: \(x\ne-2;-3;-4;-5\)
\(1+\dfrac{1}{x+2}-\left(1+\dfrac{1}{x+3}\right)=1+\dfrac{1}{x+4}-\left(1+\dfrac{1}{x+5}\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=\left(x+4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+5x+6=x^2+9x+20\)
\(\Leftrightarrow4x=-14\Rightarrow x=-\dfrac{7}{2}\)
b/ ĐK: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow x^2+3x+2-\left(x^2+5x-14\right)-12=0\)
\(\Leftrightarrow-2x+4=0\Rightarrow x=2\) (ko t/m)
Vậy pt vô nghiệm
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a)\(\dfrac{x-2}{3}-\dfrac{x-3}{4}=1\Leftrightarrow\dfrac{4x-8-3x+9}{12}=1\) ⇔x+1=12⇔x=11 Vậy phương trình đã cho có tập nghiệm S=\(\left\{11\right\}\) b)\(\dfrac{x-1}{2015}+\dfrac{x-2}{2014}+\dfrac{x-5}{2011}+\dfrac{x+1}{2017}=4\) \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+\left(\dfrac{x-5}{2011}-1\right)+\left(\dfrac{x+1}{2017}-1\right)=4-4\) \(\Leftrightarrow\dfrac{x-1-2015}{2015}+\dfrac{x-2-2014}{2014}+\dfrac{x-5-2011}{2011}+\dfrac{x+1-2017}{2017}=0\) \(\Leftrightarrow\dfrac{x-2016}{2015}+\dfrac{x-2016}{2014}+\dfrac{x-2016}{2011}+\dfrac{x-2016}{2017}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x-2016=0\) (vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\ne0\) )
⇔x=2016
Vậy phương trình đã cho có tập nghiệm S=\(\left\{2016\right\}\)
c)3(x-1)-5(x+4)+6(2-x)=7 ⇔3x-3-5x-20+12-6x=7⇔3x-5x-6x=7-12+20+3⇔-8x=18⇔\(x=\dfrac{-9}{4}\)
Vậy phương trình đã cho có tập nghiệm S=\(\left\{\dfrac{-9}{4}\right\}\)
điều kiện : \(x\ne5\)
ta có : \(\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{1}{x-5}=4\)
\(\Leftrightarrow\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{x-n}{x-5}=4\) (với \(n=x-1\))
\(\Leftrightarrow\dfrac{x-5+x-6+x-7+...+x-n}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(n-4\right)x-5\left(n-4\right)-\left(0+1+2+3+...+n-5\right)}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(x-5\right)x-5\left(x-5\right)-\left(0+1+2+3+...+x-6\right)}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(x-5\right)x-5\left(x-5\right)-\dfrac{\left(x-6\right)\left(x-5\right)}{2}}{x-5}=4\)
\(\Leftrightarrow x-5-\dfrac{x-6}{2}=4\Leftrightarrow\dfrac{2x-10-x+6}{2}=4\)
\(\Leftrightarrow\dfrac{x-4}{2}=4\Leftrightarrow x-4=2\Leftrightarrow x=6\left(tmđk\right)\)
vậy \(x=6\)
ĐK: \(x\ge5\)
\(\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{1}{x-5}=4\)
\(\Rightarrow\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{x-\left(x-1\right)}{x-5}=4\)
\(\Rightarrow\dfrac{\left(x-5\right)+\left(x-6\right)+\left(x-7\right)+...+\left[x-\left(x-1\right)\right]}{x-5}=4\) (1)
Số số hạng của tử số của vế trái của (1) là: \(x-1-5+1=x-5\)
\(\Rightarrow\dfrac{x\left(x-5\right)-\left(x-1+5\right)\left(x-5\right):2}{x-5}=4\)
\(\Rightarrow x-\left(x+4\right):2=4\)
\(\Rightarrow x-\dfrac{1}{2}x-2=4\)
\(\Rightarrow\dfrac{1}{2}x=6\)
\(\Rightarrow x=12\)