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Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.
mấy bài còn lại bạn đăng cx làm tương tự
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ....
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\\ \Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\\ \Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\\ \Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\\\Leftrightarrow x+2020=0\\\Leftrightarrow x=-2020\)
Vậy pt có tập nghiệm \(S=\left\{-2020\right\}\)
\(\Leftrightarrow\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+1+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4=0\)
\(\Leftrightarrow\left(x+2020\right)\left(...\right)=0\Rightarrow x=-2020\)
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+1+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
Mà \(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy . . . . . . . .
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4==0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
<=> x+2020=0 \(\left(\frac{1}{1996}+\frac{1}{1955}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
<=> x=-2020
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
\(\dfrac{x-1}{1992}+\dfrac{x-2}{1993}=\dfrac{x-3}{1994}+\dfrac{x-4}{1995}\)
\(\Rightarrow\left(\dfrac{x-1}{1992}+1\right)+\left(\dfrac{x-2}{1993}+1\right)=\left(\dfrac{x-3}{1994}+1\right)+\left(\dfrac{x-4}{1995}+1\right)\)
\(\Rightarrow\left(\dfrac{x-1+1992}{1992}\right)+\left(\dfrac{x-2+1993}{1993}\right)=\left(\dfrac{x-3+1994}{1994}\right)+\left(\dfrac{x-4+1995}{1995}\right)\)
\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}=\dfrac{x+1991}{1994}+\dfrac{x+1991}{1995}\)
\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}-\dfrac{x+1991}{1994}-\dfrac{x+1991}{1995}=0\)
\(\Rightarrow\left(x+1991\right)\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)=0\)
\(\Rightarrow\left(x+1991\right)=0\) ( vì \(\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)\ne0\)
\(\Rightarrow x=-1991\)
a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
=>x+36=0
=>x=-36
b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)
=>x-2004=0
=>x=2004
Lời giải:
PT đã cho tương đương với:
\(\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+1+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4=0\)
\(\Leftrightarrow \frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow (x+2020)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
Dễ thấy \(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\neq 0\) nên \(x+2020=0\Rightarrow x=-2020\) là nghiệm của pt.
Vậy............