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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
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 ................................
b) \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{18}\\< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ < =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ quyđồngmẫuvàkhửmẫu\\ x^{2^{ }}+6x-27=0\\ giảipttìmđược:x=3;x=-9\)
a) \(\frac{x-2015}{1}+\frac{x-2014}{2}+\frac{x-2013}{3}+...+\frac{x-1}{2015}+\frac{x}{2016}=0\\ \Leftrightarrow\frac{x-2015}{1}-1+\frac{x-2014}{2}-1+...+\frac{x-1}{2015}-1+\frac{x}{2016}-1=-2016\)
\(\Leftrightarrow\frac{\left(x-2016\right).1}{1}+\frac{\left(x-2016\right).1}{2}+\frac{\left(x-2016\right).1}{3}+...+\frac{\left(x-2016\right).1}{2015}+\frac{\left(x-2016\right).1}{2016}=-2016\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=-2016\)
tới đây mình chịu. mình nghĩ là phương trình bạn cho là bằng 2016 chứ, như thế giải mới được, còn như này thì mình bó tay
b)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\\ \Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x+12-32=0\\ \Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-10;2}
Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.
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=\(\left\{\dfrac{-9}{4}\right\}\)
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
\(a,\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=-3+3\)
\(\Leftrightarrow\dfrac{1+x+2017}{2017}+\dfrac{2+x+2016}{2016}+\dfrac{3+x+2015}{2015}=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b,\(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{2x+4}{5}}{15}=\dfrac{\dfrac{11x-3}{2}}{5}-\dfrac{5x-5}{5}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-\dfrac{10x-10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3-10x+10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{x+7}{10}\)
\(\Leftrightarrow10\left(2x+4\right)=75\left(x+7\right)\)
\(\Leftrightarrow20x+40=75x+525\)
\(\Leftrightarrow20x-75x=525-40\)
\(\Leftrightarrow-55x=485\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)
\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)
\(\Leftrightarrow4x+8=165x-45-150x+150\)
\(\Leftrightarrow4x-165x+150x=-45+150-8\)
\(\Leftrightarrow-11x=97\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
\(S=\left\{-\dfrac{97}{11}\right\}\)