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a ) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}+\dfrac{x+4}{97}=-4\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}+\dfrac{x+101}{97}=-1\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{97}\right)=-1\)
Vô lí => Phương trình trên vô nghiệm .
b ) \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-73}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89.
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-12}{77}+\dfrac{x-11}{78}-\dfrac{x-74}{15}-\dfrac{x-73}{16}=0\)
\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1-\dfrac{x-74}{15}+1-\dfrac{x-73}{16}+1=0+1+1-1-1\)
\(\Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)-\left(\dfrac{x-74}{15}-1\right)-\left(\dfrac{x-73}{16}-1\right)=0\)
\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-89=0\\\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\end{matrix}\right.\)
\(x-89=0\\ \Rightarrow x=89\)
\(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\)(vô lí)
Vậy \(x=89\)
a)\(\dfrac{1}{2}\)(x+1)+\(\dfrac{1}{4}\)(x+3)=3-\(\dfrac{1}{3}\)(x+2)
\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)x+\(\dfrac{3}{4}\)=3-\(\dfrac{1}{3}\)x-\(\dfrac{2}{3}\)
\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{4}\)x+\(\dfrac{1}{3}\)x=-\(\dfrac{1}{2}\)-\(\dfrac{3}{4}\)+3-\(\dfrac{2}{3}\)
\(\Leftrightarrow\)\(\dfrac{13}{12}\)x=\(\dfrac{13}{12}\)
\(\Leftrightarrow\)x=1
Vậy nghiệm của pt là x=1
b)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)=\(\dfrac{x+6}{94}\)+\(\dfrac{x+8}{92}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)-\(\dfrac{x+6}{94}\)-\(\dfrac{x+8}{92}\)=0
\(\Leftrightarrow\)(\(\dfrac{x+2}{98}\)+1)+(\(\dfrac{x+4}{96}\)+1)-(\(\dfrac{x+6}{94}\)+1)-(\(\dfrac{x+8}{92}\)+1)=0
\(\Leftrightarrow\)\(\dfrac{x+2+98}{98}\)+\(\dfrac{x+4+96}{96}\)-\(\dfrac{x+6+94}{94}\)-\(\dfrac{x+8+92}{92}\)=0
\(\Leftrightarrow\)\(\dfrac{x+100}{98}\)+\(\dfrac{x+100}{96}\)-\(\dfrac{x+100}{94}\)-\(\dfrac{x+100}{92}\)=0
\(\Leftrightarrow\)(x+100)(\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\))=0
\(\Leftrightarrow\)x+100=0(vì\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\)\(\ne\)0)
\(\Leftrightarrow\)x=-100
Vậy nghiệm của pt là x=-100
a) \(\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\)
\(\Rightarrow\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+4=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+1\right)+\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+2}{2013}+1\right)=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+\dfrac{2010}{2010}\right)+\left(\dfrac{x+4}{2011}+\dfrac{2011}{2011}\right)+\left(\dfrac{x+3}{2012}+\dfrac{2012}{2012}\right)+\left(\dfrac{x+2}{2013}+\dfrac{2013}{2013}\right)=0\)
\(\Rightarrow\dfrac{x+5+2010}{2010}+\dfrac{x+4+2011}{2011}+\dfrac{x+3+2012}{2012}+\dfrac{x+2+2013}{2013}=0\)
\(\Rightarrow\dfrac{x+2015}{2010}+\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2013}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
b) \(\dfrac{x-22}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-22}{77}+\dfrac{x-11}{78}-2=\dfrac{x-74}{15}+\dfrac{x-73}{16}-2\)
\(\Rightarrow\left(\dfrac{x-22}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\)
\(\Rightarrow\left(\dfrac{x-22}{77}-\dfrac{77}{77}\right)+\left(\dfrac{x-11}{78}-\dfrac{78}{78}\right)=\left(\dfrac{x-74}{15}-\dfrac{15}{15}\right)+\left(\dfrac{x-73}{16}-\dfrac{16}{16}\right)\)
\(\Rightarrow\dfrac{x-22-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)
\(\Rightarrow\dfrac{x-99}{77}+\dfrac{x-99}{78}=\dfrac{x-99}{15}+\dfrac{x-99}{16}\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)=\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)-\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
Mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)
\(\Rightarrow x-99=0\)
\(\Rightarrow x=99\)
a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
Giải các bất phương trình sau :
a) \(\left(x-1\right)\left(x+3\right)< 0\)
Lập bảng xét dấu :
x x-1 x+3 (x-1)(x+3) -3 1 - 0 + - 0 - + + + - +
Nghiệm của bất phương trình là : \(-3< x< 1\)
b) \(\left(2x-1\right)\left(x+2\right)>0\)
Lập bảng xét dấu :
x 2x-1 x+2 (2x-1)(x+2) -2 1 2 0 0 - - + - + + - + +
Nghiệm của bất phương trình là : \(x< -2;x>\dfrac{1}{2}\)
c) \(\dfrac{3x-2}{2x-1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-2\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x< \dfrac{1}{2}\end{matrix}\right.\)
d) \(\dfrac{3x+2}{x+1}>2\)
\(\Leftrightarrow\dfrac{3x+2}{x+1}-\dfrac{2\left(x+1\right)}{x+1}>0\)
\(\Leftrightarrow\dfrac{3x+2-2x-2}{x+1}>0\)
\(\Leftrightarrow\dfrac{x}{x+1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}x\ge0\\x< -1\end{matrix}\right.\)
a, (x-1)(x+3) <0
TH1: x-1<0<=>x<1
x+3>0<=>x>-3
=>-3<x<1
TH2: x-1>0<=>x>1
x+3<0<=>x<-3
=>Vô lý
Vậy S={x|-3<x<1}
b,(2x-1)(x+2)>0
TH1: 2x-1\(\ge\)0<=>2x\(\ge\)1<=>x\(\ge\)\(\dfrac{1}{2}\)
x+2\(\ge\)0<=>x\(\ge\)-2
=>x\(\ge\)\(\dfrac{1}{2}\)
TH2: 2x-1<0<=>2x<1<=>x<\(\dfrac{1}{2}\)
x+2<0<=>x<-2
=>x<-2
Vậy S={x|x<-2 hoặc x\(\ge\)\(\dfrac{1}{2}\)}
c, \(\dfrac{3x-2}{2x-1}\)>0 (Tử và mẫu cùng dấu)
TH1 3x-2\(\ge\)0<=>3x\(\ge\)2<=>x\(\ge\)2
2x-1>0<=>2x>1<=>x>\(\dfrac{1}{2}\)
=>x\(\ge\)2
TH2: 3x-2<0<=>3x<2<=>x<\(\dfrac{2}{3}\)
2x-1<0<=>2x<1<=>x<\(\dfrac{1}{2}\)
=>x<\(\dfrac{1}{2}\)
Vậy S={x|x\(\ge\)2 hoặc x<\(\dfrac{1}{2}\)}
d,\(\dfrac{3x+2}{x+1}>2\)
<=>\(\dfrac{3x+2}{x+1}-2\)>0
<=>\(\dfrac{3x-2-2x-2}{x+1}\)>0
<=>\(\dfrac{x-4}{x+1}\)>0 (Tử và mẫu cùng dấu)
TH1: x-4\(\ge\)0<=>x\(\ge\)4
x+1>0<=>x>-1
=>x\(\ge\)-4
TH2: x-4<0<=>x<4
x+1<0<=>x<-1
=>x<-1
Vậy S={x|x\(\ge\)-4 hoặc x<-1}
easy làm câu b vs c trước nha
b) \(\left(x-5\right)\left(2x+4\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\2x+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\2x+4< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\)
Vậy......
c) \(\left(x+3\right)\left(3x-6\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\3x-6< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\3x-6>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3< x< 2\\x\in\varnothing\end{matrix}\right.\)
Vậy.......
- Vậy còn câu a ?