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Giải phương trình sau:
\(\frac{x-2004}{15}\)+\(\frac{x-1995}{12}\)+\(\frac{x-1989}{10}\)+\(\frac{x-1987}{8}\)=\(10\)
⇔\(\frac{\left(x-2004\right).40}{600}\) +\(\frac{\left(x-1995\right).50}{600}\)+\(\frac{\left(x-1989\right).60}{600}\)+\(\frac{\left(x-1987\right).75}{600}\)=\(\frac{10.600}{600}\)
⇔\(\frac{40x-80160}{600}\) + \(\frac{50x-99750}{600}\) +\(\frac{60x-119340}{600}\) +\(\frac{75x-149025}{600}\)=\(\frac{6000}{600}\)
➞ \(40x-80160+50x-99750+60x-119340+75x-149025=6000\)⇔\(225x=\)\(6000+80160+99750+119340+149025\)
⇔\(225x=454275\)
⇔\(x=2019\)
\(\frac{x+6}{1999}+\frac{x+8}{1997}=\frac{x+10}{1995}+\frac{x+12}{1993}\)
\(\Leftrightarrow\frac{x+6}{1999}+1+\frac{x+8}{1997}+1=\frac{x+10}{1995}+1+\frac{x+12}{1993}+1\)
\(\Leftrightarrow\frac{x+2005}{1999}+\frac{x+2005}{1997}=\frac{x+2005}{1995}+\frac{x+2005}{1993}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2005=0\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\ne0\right)\)
<=> 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)
Giải phương trình sau:
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4.\)-4
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4\)
\(\Rightarrow\frac{x+2001}{5}+1+\frac{x+1999}{7}+1+\frac{x+1997}{9}+1+\frac{x+1995}{11}+1=0\)
\(\Rightarrow\frac{x+2006}{5}+\frac{x+2006}{7}+\frac{x+2006}{9}+\frac{x+2006}{11}=0\)
\(\Rightarrow\left(x+2006\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)=0\)
\(\Rightarrow x+2006=0\)vì \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}>0\)
\(\Rightarrow x=-2006\)
Bài làm
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x+2+2005}{2005}\right)+\left(\frac{x+3+2004}{2004}\right)+\left(\frac{x+4+2003}{2003}\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right).\frac{1}{2005}+\left(x+2007\right).\frac{1}{2004}+\left(x+2007\right).\frac{1}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2007=\frac{0}{\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}}\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy phương trình trên có tập nghiệm S = { -2007 }
# Học tốt #
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}>0\)(2)
Từ (1), (2) \(\Rightarrow x+2017=0\)\(\Leftrightarrow x=-2017\)
Vậy \(x=-2017\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
\(\frac{15x-10}{x^2+3}=0\)
\(\Leftrightarrow\frac{5\left(3x-2\right)}{x^2+3}=0\)
\(\Leftrightarrow5\left(3x-2\right)=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
...
x mzn cjk
nguyễnđìnhquyết 2k9
ko bt thì đừng có đăng linh tinh!