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\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))
\(\Leftrightarrow x=-2004\)
\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)
\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)
\(\Leftrightarrow x=-2004\)
Vậy PT có tập nghiệm S = {-2004}
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000
\(\frac{x+6}{1999}+\frac{x+8}{1997}=\frac{x+10}{1995}+\frac{x+12}{1993}\)
\(\Leftrightarrow\frac{x+6}{1999}+1+\frac{x+8}{1997}+1=\frac{x+10}{1995}+1+\frac{x+12}{1993}+1\)
\(\Leftrightarrow\frac{x+2005}{1999}+\frac{x+2005}{1997}=\frac{x+2005}{1995}+\frac{x+2005}{1993}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2005=0\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\ne0\right)\)
<=> x=-2005
Vậy x=-2005
bạn chỉ cần cộng mỗi phân số với 1 là xong!
Vd: x+6/1999 +1 +x+8/1997 +1 = x+10/1995 +1 +x+12/1993 +1
(không quen sử dụng cái phần mềm này lắm nên mình không làm nốt được)
a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)
=>x+2006=0
=>x=-2006
b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)
=>x-105=0
=>x=105
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
<=> \(\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)
<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0
<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)
<=> x = 2004
Vậy S = {2004}
đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)
\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)
=> x=1
cái cuối là =-4 nhé!
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4\)
\(\Rightarrow\frac{x+2001}{5}+1+\frac{x+1999}{7}+1+\frac{x+1997}{9}+1+\frac{x+1995}{11}+1=0\)
\(\Rightarrow\frac{x+2006}{5}+\frac{x+2006}{7}+\frac{x+2006}{9}+\frac{x+2006}{11}=0\)
\(\Rightarrow\left(x+2006\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)=0\)
\(\Rightarrow x+2006=0\)vì \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}>0\)
\(\Rightarrow x=-2006\)