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Tham khảo bài 4 của bạn ở đây (Cách làm và kết quả)
Vũ Minh TuấnNguyễn Ngọc Lộc Phạm Thị Diệu HuyềnPhạm Minh QuangTrên con đường thành công không có dấu chân của kẻ lười biếngHISINOMA KINIMADOTrần Thanh PhươngNguyễn Lê Phước Thịnhbach nhac lam
\(a.\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\\ \Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\\\Leftrightarrow \left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\\\Leftrightarrow x-23=0\left(Vi\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\right)\\\Leftrightarrow x=23\)
\(b.\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\\\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\\ \Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\\ \Leftrightarrow x+100=0\left(Vi\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\\ \Leftrightarrow x=-100\)
\(c.\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\ \Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\\\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\\ \Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\\\Leftrightarrow x+2005=0\left(Vi\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\right)\\ \Leftrightarrow x=-2005\)
\(d.\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\\\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\\\Leftrightarrow \left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\\ \Leftrightarrow300-x=0\left(vi\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\right)\\ \Leftrightarrow x=300\)
\(e.\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\\ \Leftrightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\\ \Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\\ \Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\\ \Leftrightarrow x-100=0\left(vi\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\right)\\\Leftrightarrow x=100\)
\(f.\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\\ \Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\\ \Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\\ \Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\\ \Leftrightarrow x+10=0\left(vi\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\right)\\ \Leftrightarrow x=-10\)
\(h.\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\\\Leftrightarrow \frac{2-x}{2002}+1=\frac{1-x}{2003}+1+\frac{-x}{2004}+1\\\Leftrightarrow\frac{2004-x}{2002}-\frac{2004-x}{2003}-\frac{2004-x}{2004}=0\\\Leftrightarrow \left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(vi\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\right)\\\Leftrightarrow x=2004\)
\(g.\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\\\Leftrightarrow \frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\\\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\\\Leftrightarrow \left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\\\Leftrightarrow x+100=0\left(vi\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\\\Leftrightarrow x=-100\)
Nốt câu i bài 4 của bạn nha
\(i.\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-71}{1929}+\frac{x^2-10x-73}{1927}\\ \Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-71}{1929}-1+\frac{x^2-10x-73}{1923}-1\\\Leftrightarrow \frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{1929}-\frac{x^2-10x-2000}{1927}=0\\\Leftrightarrow \left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{1929}-\frac{1}{1927}\right)=0\)
\(\Leftrightarrow x^2-10x-2000=0\left(Vi\frac{1}{1971}+\frac{1}{1973}-\frac{1}{1929}-\frac{1}{1927}\ne0\right)\\\Leftrightarrow x^2-50x+40x-2000=0\\\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\\ \Leftrightarrow\left(x+40\right)\left(x-50\right)=0\\\Rightarrow \left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-40;50\right\}\)
Cho 3 điểm A, B, C theo thứ tự nằm trên đường thẳng d, ( AB > BC), Trên cùng 1 nửa mặt phẳng bờ là đường thẳng d, vẽ các 
đều, Gọi M, N, P, Q, I theo thứ tự là Trung điểm của các đoạn thẳng BD, AE, BE, CD, DE
a, CMR: 3 điểm I, M, N thẳng hàng b, CMR: 3 điểm I, Q, P thẳng hàng
c, CMR: MNPQ là thình thang cân d, 
NQ = \(\frac{1}{2}\) DE
