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a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)
\(\Rightarrow x+10=0\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)
\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0
Vậy x = 2012
a, (x+1)/9 +1 + (x+2)/8 = (x+3)/7 + 1 + (x+4)/6 + 1
<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6
<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0
vì 1/9 +1/8 -1/7 - 1/6 khác 0
=> x+10=0
=> x=-10
a, \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
= \(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-1}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
( x + 5)(x - 3) = \(x^2-1\) - 8
x\(^2\) -3x + 5x -15 = \(x^2-9\)
= > \(x^2-x^2\) +2x = 15 - 9
=> 2x = 6
=> x = 3
a, Ta có : \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=-5+5=0\)
=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
=> \(424-x=0\)
=> \(x=424\)
Vậy phương trình có nghiệm là x = 424 .
b, Ta có : \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
=> \(\frac{x+1}{2014}+1+\frac{x+3}{2012}+1=\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}-\frac{x+2015}{2010}-\frac{x+2015}{2009}=0\)
=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> \(x+2015=0\)
=> \(x=-2015\)
Vậy phương trình có nghiệm là x = -2015 .
a) \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
<=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{40}\right)=0\)
<=> 424 - x = 0
<=> x = 424
Vậy S = {424}
b) \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
<=> x + 2015 = 0
<=> x= -2015
Vậy S = {-2015}
\(\Leftrightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Leftrightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x=2012\)
a) \(=-7\left(x^2-\frac{10}{7}x+\frac{2016}{7}\right)\)
\(=-7\left(x^2-2.\frac{5}{7}x+\frac{25}{49}+\frac{14087}{49}\right)\)
\(=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\)
ta có
\(\left(x-\frac{5}{7}\right)^2\ge0\)với mọi x
\(=>-7\left(x-\frac{5}{7}\right)^2\le0\)(nhân cả hai vế với -7)
\(=>-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)
trường hợp dấu "=" xảy ra khi và chỉ khi
\(\left(x-\frac{5}{7}\right)^2=0\)
\(=>x-\frac{5}{7}=0\)
\(=>x=\frac{5}{7}\)
vậy GTLN cảu biểu thức là \(-\frac{14087}{7}\) khi và chỉ khi x= \(\frac{5}{7}\)
a) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-1}{17}-5=0\)
\(\Leftrightarrow\frac{x-90-10}{10}+\frac{x-76-2.12}{12}+\frac{x-58-3.14}{14}+\frac{x-36-4.16}{16}+\frac{x-15-5.17}{17}=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b) \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Leftrightarrow\frac{x+2011}{2013}+1+\frac{x+2012}{2012}+1=\frac{x+2010}{2014}+1+\frac{x+2013}{2011}+1\)
\(\Leftrightarrow\frac{x+2011+2013}{2013}+\frac{x+2012+2012}{2012}=\frac{x+2010+2014}{2014}+\frac{x+2013+2011}{2011}\)
\(\Leftrightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Leftrightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Leftrightarrow x+4024=0\Leftrightarrow x=-4024\)
Vậy \(S=\left\{-4024\right\}\)
Phương trình a bạn trừ phân thức đầu tiên cho 1, phân thức thứ hai cho 2, phân thức thứ ba cho 3, phân thức thứ tư cho 4, phân thức thứ năm cho 5, vế còn lại trừ đi 15. Tiếp theo bạn đặt x -100 làm nhân tử chung. Cuối cùng tìm được x= 100
Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+....+\frac{x-2012}{1}-1=2012-2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2011}+....+1\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=2013\)
Vậy x = 2013
PT đã cho tương đương với:
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2010=2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
\(\Leftrightarrow x=2013\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
tự làm nốt~
kudo shinichi làm sai ở chỗ:
\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé
. Ta có: \(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\) \(\Leftrightarrow\frac{x+1}{2016}+1+\frac{x+3}{2014}+1=\frac{x+5}{2012}+1\frac{x+7}{2010}+1\)
. \(\Leftrightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\) \(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)\)
\(\Leftrightarrow x+2017=0\) \(\Leftrightarrow x=-2017\)
\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2016}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+7}{2010}+1\right)\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}=\frac{x+2017}{2012}+\frac{x+2017}{2010}\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2017=0\)\(\left(Vì\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\ne0\right)\)
\(\Rightarrow x=0-2017\)
\(\Rightarrow x=-2017\)
Vậy x=-2017