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\(\frac{x-30}{10}+\frac{x-28}{9}+\frac{x-26}{8}=-6\)
<=> \(\frac{36.\left(x-30\right)}{360}+\frac{40\left(x-28\right)}{360}+\frac{45\left(x-26\right)}{360}=\frac{-2160}{360}\)
=> \(36x-1080+40x-1120+45x-1170=-2160\)
\(< =>36x+40x+45x=-2160+1080+1120+1170\)
<=> \(121x=1210\)
<=> x = 10
\(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(=>\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(=>\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(=>\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
Do \(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\ne0\)
\(=>x-30=0\)
\(=>x=30\)
Vậy nghiệm của phương trình trên là 30
Học tốt
\(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\Leftrightarrow\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\Leftrightarrow\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\Leftrightarrow\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
Vì \(\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)>0\)
\(\Rightarrow x-30=0\)
\(\Rightarrow x=30\)
\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\text{ma}:\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
=> x + 0 = 10
=> x = 0 -10
=> x = -10
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)<=> \(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)
<=>\(\frac{8+x-8}{x-8}+\frac{11+x-11}{x-11}=\frac{9+x-9}{x-9}+\frac{10+x-10}{x-10}\)
<=>\(\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)
<=>\(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=>\(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x-8}+\frac{1}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\end{cases}}\)
đến đoạn bạn giải tiếp nhé
a/ ĐKXĐ: \(x\ne\left\{8;9;10;11\right\}\)
\(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)
\(\Leftrightarrow\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)
\(\Leftrightarrow x\left(\frac{1}{x-8}-\frac{1}{x-9}+\frac{1}{x-11}-\frac{1}{x-10}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-9}-\frac{1}{x-8}=\frac{1}{x-11}-\frac{1}{x-10}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{1}{\left(x-9\right)\left(x-8\right)}=\frac{1}{\left(x-11\right)\left(x-10\right)}\)
\(\Leftrightarrow x^2-17x+72=x^2-21x+110\)
\(\Rightarrow x=\frac{19}{2}\)
b/ ĐK: \(x\ne\left\{3;4;5;6\right\}\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\frac{-2}{\left(x-3\right)\left(x-5\right)}=\frac{-2}{\left(x-4\right)\left(x-6\right)}\)
\(\Leftrightarrow x^2-8x+15=x^2-10x+24\)
\(\Rightarrow x=\frac{9}{2}\)
a, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(1)
ĐKXĐ: \(\hept{\begin{cases}x+9\ne0\\x+10\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-9\\x\ne-10\end{cases}}}\)
(1)\(\Leftrightarrow\frac{9.\left(x+9\right)}{90}+\frac{10.\left(x+10\right)}{90}=\frac{9.\left(x+9\right)}{\left(x+9\right)\left(x+10\right)}+\frac{10.\left(x+10\right)}{\left(x+9\right)\left(x+10\right)}\)
\(\Leftrightarrow9.\left(x+9\right)+10.\left(x+10\right)=9.\left(x+9\right)+10.\left(x+10\right)\)
\(\Leftrightarrow9x+81+10x+100=9x+81+10x+100\)
\(\Leftrightarrow9x+10x-9x-10x=81+100-81-100\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow x\in R\)trừ -9 và -10
a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có nghiệm là x = 2004 .
b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)
=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình có nghiệm là x = 100 .
thanks
Từ đề bài, ta có:
\(2+\frac{x-30}{10}+2+\frac{x-28}{9}+2+\frac{x-26}{8}=0\)
\(\Leftrightarrow\frac{x-10}{10}+\frac{x-10}{9}+\frac{x-10}{8}=0\)
\(\Leftrightarrow\left(x-10\right)\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}\right)=0\)
Do \(\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}\right)>0\)nên x-10=0
<=> x=10
Vậy phương trình có nghiệm duy nhất x=10
@Gà Mờ Thks bạn