Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
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
a) Đặt x -3 = a
<=> a(a+2)(a+8)(a+10) - 297=0
<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0
<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)
Đặt \(a^2+10a=b\)
\(b^2+16b-297=0\)
\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)
b) bấm máy ra nhân tử chung :D
c)
\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)
\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
<=> x = 2018
d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)
giống câu c
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 ................................
1a)\(\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
=> \(\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
=>\(\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
=> x=100( vi \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\)
b) \(\dfrac{x-5}{2012}-1+\dfrac{x-4}{2013}-1=\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1\)
=> \(\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
=>(x-2017).\(\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
=> x=2015(vi .....................................................≠0)
2)
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
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}
a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100