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E
29 tháng 3 2017
BPT \(\Leftrightarrow\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}>0\)
\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)>0\)
\(\Leftrightarrow\dfrac{x-15}{2002}+\dfrac{x-15}{2003}-\dfrac{x-15}{2004}-\dfrac{x-15}{2005}>0\)
\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)
Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}>0\)
\(\Rightarrow x-15>0\)
\(\Leftrightarrow x>15\)
Vậy bpt có nghiệm x > 15
TK
29 tháng 3 2017
\(\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-2>\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}-2\)
\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)\)
\(-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)\)
quy đồng lên ta được:
\(\left(\dfrac{x+1987-2002}{2002}\right)+\left(\dfrac{x-1998-2003}{2003}\right)\)
\(-\left(\dfrac{x+1989-2004}{2004}\right)-\left(\dfrac{x+1990-2005}{2005}\right)>0\)
\(\Leftrightarrow\left(\dfrac{x-15}{2002}\right)+\left(\dfrac{x-15}{2003}\right)-\left(\dfrac{x-15}{2004}\right)-\left(\dfrac{x-15}{2005}\right)>0\)
đặt nhân tử chung ta được:
\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)
Vì:
\(\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\in Z\right)\) nên ta xét \(x-15>0\Rightarrow x>15\)
21 tháng 1 2018
a) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
b) \(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}=\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+8}{1996}+1=\dfrac{x+12}{1992}+1+\dfrac{x+16}{1988}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{1996}-\dfrac{x+2004}{1992}-\dfrac{x+2004}{1988}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì \(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\ne0\))
\(\Leftrightarrow x=-2004\)
25 tháng 6 2020
\(\frac{x-21}{1999}+\frac{x-33}{1987}\le\frac{x+6}{2026}+\frac{x+11}{2031}\)
<=> \(\frac{x-21}{1999}-1+\frac{x-33}{1987}-1\le\frac{x+6}{2026}-1+\frac{x+11}{2031}-1\)
<,=>. \(\frac{x-2020}{1999}+\frac{x-2020}{1987}\le\frac{x-2020}{2026}+\frac{x-2020}{2031}\)
<=> \(\left(x-2020\right)\left(\frac{1}{1999}+\frac{1}{1987}-\frac{1}{2026}-\frac{1}{2031}\right)\le0\) (1)
Vì \(\frac{1}{1999}+\frac{1}{1987}-\frac{1}{2026}-\frac{1}{2031}\ge0\)
Nên (1) \(x-2020\le0\Leftrightarrow x\le2020\)
5 tháng 3 2015
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
5 tháng 8 2016
a) đề bài => \(\frac{159-x}{141}+1+\frac{157-x}{143}+1+...+\frac{151-x}{149}+1=0\)
=>\(\frac{300-x}{141}+\frac{300-x}{143}+...+\frac{300-x}{149}=0\)
=>\(\left(300-x\right).\left(\frac{1}{141}+\frac{1}{143}+...+\frac{1}{149}\right)=0\)
vì \(\frac{1}{141}+\frac{1}{143}+...+\frac{1}{149}\ne0\)
=> \(300-x=0\)
=>\(x=300\)
chờ mình chút sẽ có câu b. k cho mình nha.