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\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\\ \dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>-3+3\\ \dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\\ \left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\\ x+2954>0\\ x>-2954\)
a.2mx=0 <=> mx=0
•nếu m=0 thì nghiệm đúng với mọi x
•nếu \(m\ne0\) thì nghiệm đúng với x=0
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}>-2\)
\(\Leftrightarrow\dfrac{x+1}{2953}+1+\dfrac{x+953}{2001}+1>-2+1+1\)
\(\Leftrightarrow\dfrac{x+2954}{2953}+\dfrac{x+2955}{2001}>0\)
\(\Leftrightarrow\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}\right)>0\)
\(\Leftrightarrow x+2954>0\\ \Leftrightarrow x>-2954\)
Vậy.......
a: \(x>3:\dfrac{1}{2}=6\)
b: \(x>-2:\left(-\dfrac{1}{3}\right)=6\)
c: \(x>-4:\dfrac{2}{3}=-6\)
d: \(x< -6:\dfrac{3}{5}=-10\)
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x+\left(2x-1\right)}{6}=\dfrac{24-2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x+2x=24+1\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{25}{8}\)
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)+3\left(x-1\right)}{12}=\dfrac{12-8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow9\left(x-1\right)+8\left(x-1\right)=12\)
\(\Leftrightarrow17\left(x-1\right)=12\)
\(\Leftrightarrow17x-17=12\)
\(17x=12+17\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{29}{17}\)
c) \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-\dfrac{1-x}{2002}-\dfrac{\left(-x\right)}{2003}=1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1-\dfrac{1-x}{2002}-1-\dfrac{\left(-x\right)}{2003}-1=1+1-1-1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+\dfrac{2001}{2001}-\dfrac{1-x}{2002}-\dfrac{2002}{2002}-\dfrac{\left(-x\right)}{2003}-\dfrac{2003}{2003}=0\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\)
\(\Leftrightarrow-x=-2003\)
\(\Leftrightarrow x=2003\)
Vậy phương trình có một nghiệm là x = 2003
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}=\dfrac{24}{6}-\dfrac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow4x+2x+2x=1+24\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy S={\(\dfrac{25}{8}\)}
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}+\dfrac{3\left(x-1\right)}{12}=\dfrac{12}{12}-\dfrac{8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow6x-6+3x-3=12-8x+8\)
\(\Leftrightarrow6x+3x+8x=6+3+12+8\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy S={\(\dfrac{29}{17}\)}
a) \(\dfrac{1-2x}{4}-2< \dfrac{1-5x}{8}\\ < =>\dfrac{2-4x}{8}-\dfrac{16}{8}< \dfrac{1-5x}{8}\\ < =>2-4x-16< 1-5x\\ < =>-4x+5x< 1-2+16\\ < =>x< 15\)
Vậy : tập nghiệm của bất phương trình là S= \(\left\{x|x< 15\right\}\)
b) \(\dfrac{x-1}{4}-1>\dfrac{x+1}{3}+8\\ < =>\dfrac{3x-3}{12}-\dfrac{12}{12}>\dfrac{4x+4}{12}+\dfrac{96}{12}\\ < =>3x-3-12>4x+4+96\\ < =>3x-4x>4+96+3+12\\ < =>-x>115\\ =>x< -115\)
Vậy: tập nghiệm của bất phương trình là S=\(\left\{x|x< -115\right\}\)
a: \(x< -9:\dfrac{3}{2}=-9\cdot\dfrac{2}{3}=-6\)
b: 2/3x>-2
hay x>-2:2/3=-3
c: \(2x>\dfrac{9}{5}-\dfrac{4}{5}=1\)
hay x>1/2
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}>6-4=2\)
hay x>2:3/5=2x5/3=10/3
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
Giải:
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\)
\(\Leftrightarrow\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>0\)
\(\Leftrightarrow\dfrac{x+1}{2953}+1+\dfrac{x+953}{2001}+1+\dfrac{x+2950}{4}+1>0\)
\(\Leftrightarrow\dfrac{x+1+2953}{2953}+\dfrac{x+953+2001}{2001}+\dfrac{x+2950+4}{4}>0\)
\(\Leftrightarrow\dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\)
\(\Leftrightarrow\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\)
Vì \(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}>0\)
Nên \(x+2954>0\)
\(\Leftrightarrow x>-2954\)
Vậy ...
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>3\)
<=>\(\left(\dfrac{x+1}{2953}+1\right)+\left(\dfrac{x+953}{2001}+1\right)+\left(\dfrac{x+2950}{4}+1\right)>0\)
<=>\(\dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\)
<=>\(\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\)
Vì \(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}>0\) nên \(x+2954>0\) <=> \(x>-2954\)
KL: ...