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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}
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\) \(\Leftrightarrow\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}-\dfrac{x-2017}{2017}\) \(\Leftrightarrow\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
Vì: \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
Suy ra: x -2017 = 0
=> x = 2017
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{2008}-2+\dfrac{x}{2017}-1\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2008}+\dfrac{x-2017}{2017}\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2008}-\dfrac{x-2017}{2017}=0\)
⇔\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2008}-\dfrac{1}{2017}\right)=0\)
⇔x-2017=0
⇔x=2017
vậy phương trình có tập nghiệm là S={2017}
a) \(\frac{6-x}{3}-\frac{x}{4}=\frac{3+2x}{2}-1\)
\(\frac{4\left(6-x\right)}{12}-\frac{3x}{12}=\frac{3+2x}{2}-\frac{2}{2}\)
\(\frac{24-4x-3x}{12}=\frac{3+2x-2}{2}\)
\(\frac{24-7x}{12}=\frac{2x+1}{2}\)
\(\Rightarrow2\left(24-7x\right)=12\left(2x+1\right)\)
\(\Rightarrow48-14x=24x+12\)
\(\Rightarrow24x+14x=48-12\)
\(\Rightarrow38x=36\)
\(\Rightarrow x=\frac{18}{19}\)
b) \(-7x-\frac{x-3}{5}-\frac{x}{2}=x+\frac{2x+1}{3}\)
\(\frac{-70x}{10}-\frac{2\left(x-3\right)}{10}-\frac{5x}{10}=\frac{3x}{3}+\frac{2x+1}{3}\)
\(\frac{-70x-2x+6-5x}{10}=\frac{3x+2x+1}{3}\)
\(\frac{-77x+6}{10}=\frac{5x+1}{3}\)
\(\Rightarrow3\left(-77x+6\right)=10\left(5x+1\right)\)
\(\Leftrightarrow-231x+18=50x+10\)
\(\Leftrightarrow50x+231x=18-10\)
\(\Leftrightarrow281x=8\)
\(\Leftrightarrow x=\frac{8}{281}\)
Mấy câu kia tương tự
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=\(\left\{\dfrac{-9}{4}\right\}\)
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
Giải bpt sau:
\(\dfrac{x+3}{2011}\)+\(\dfrac{x+2}{2012}\)+\(\dfrac{x+1}{2013}\)≥\(\dfrac{3x}{2014}\)
\(\dfrac{x+3}{2011}+\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\ge\dfrac{3x}{2014}\)
\(\dfrac{x+3}{2011}+1+\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\ge\dfrac{3x}{2014}+3\)
\(\dfrac{x+2014}{2011}+\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\ge3\left(\dfrac{x+2014}{2014}\right)\)
\(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)\ge0\)
Mà \(\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)>0\) (bạn có thể chứng minh nếu thích )
Nên \(x+2014\ge0\)
\(\Leftrightarrow x\ge-2014\)
Vậy
\(\dfrac{1-x}{2013}=1+\dfrac{2-x}{2012}-\dfrac{x}{2014}\)
\(\Leftrightarrow1+\dfrac{1-x}{2013}=1+\dfrac{2-x}{2013}+1-\dfrac{x}{2014}\)
\(\Leftrightarrow\dfrac{2013+1-x}{2013}=\dfrac{2012+2-x}{2012}+\dfrac{2014-x}{2014}\)
\(\Leftrightarrow\dfrac{2014-x}{2013}=\dfrac{2014-x}{2012}+\dfrac{2014-x}{2014}\)
\(\Leftrightarrow\dfrac{2014-x}{2013}-\dfrac{2014-x}{2012}-\dfrac{2014-x}{2014}=0\)
\(\Leftrightarrow\left(2014-x\right)\left(\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2014}\right)=0\)
\(\Leftrightarrow2014-x=0\) ( Vì \(\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2014}\ne0\) )
\(\Leftrightarrow x=2014\)
Vậy pt có nghiệm x = 2014
\(\dfrac{1-x}{2013}=1+\dfrac{2-x}{2012}-\dfrac{x}{2014}\)
\(\Leftrightarrow\dfrac{1-x}{2013}=\dfrac{2-x}{2012}+\dfrac{2014-x}{2014}\)
\(\Leftrightarrow\dfrac{1-x}{2013}+1=\dfrac{2-x}{2012}+1+\dfrac{2014-x}{2014}\)
\(\Leftrightarrow\dfrac{2014-x}{2013}=\dfrac{2014-x}{2012}+\dfrac{2014-x}{2014}\)
\(\Leftrightarrow\left(2014-x\right)\left(\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow2014-x>0\) (Vì \(\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Leftrightarrow x=2014\)
Vậy pt có tập nghiệm là x = 2014
\(\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+\dfrac{x+16}{4}=4\)
\(\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}=4-\dfrac{x+16}{4}\)
\(\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}\right)=-x\)
Mk giải đế đây rùi bạn tự giải nốt đi
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
Sửa đề: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
Ta có: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
\(\Leftrightarrow\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}+1=\dfrac{x+2001}{15}+1+\dfrac{x+2014}{2}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}=\dfrac{x+2016}{15}+\dfrac{x+2016}{2}\)
\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}-\dfrac{x+2016}{15}-\dfrac{x+2016}{2}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\ne0\)
nên x+2016=0
hay x=-2016
Vậy: S={-2016}