Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)<=> \(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)
<=>\(\frac{8+x-8}{x-8}+\frac{11+x-11}{x-11}=\frac{9+x-9}{x-9}+\frac{10+x-10}{x-10}\)
<=>\(\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)
<=>\(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=>\(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x-8}+\frac{1}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\end{cases}}\)
đến đoạn bạn giải tiếp nhé
=>\(\frac{30x\left(x-6\right)}{x\left(x+10\right)\left(x-6\right)}+\frac{30x\left(x+10\right)}{x\left(x+10\right)\left(x-6\right)}=\frac{60\left(x+10\right)\left(x-6\right)}{x\left(x-6\left(x+10\right)\right)}\)
=>30x2-180x+30x2+300x=60x2-360x+600x-3600
=>60x2+120x=60x2+240x-3600
=>-120x=-3600
=>x=30
nhớ k mk........Đúng 100%
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: \(x\ne\left\{3;4;5;6\right\}\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\frac{-2}{\left(x-3\right)\left(x-5\right)}=\frac{-2}{\left(x-4\right)\left(x-6\right)}\)
\(\Leftrightarrow x^2-8x+15=x^2-10x+24\)
\(\Rightarrow x=\frac{9}{2}\)
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(-537x^2+5054x=-541x^2+5092x\)
\(-537x^2+5054x+541x^2-5092x=0\)
\(4x^2-38x=0\)
\(x\left(2x-19\right)=0\)
\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)
\(\frac{x+6}{1999}+\frac{x+8}{1997}=\frac{x+10}{1995}+\frac{x+12}{1993}\)
\(\Leftrightarrow\frac{x+6}{1999}+1+\frac{x+8}{1997}+1=\frac{x+10}{1995}+1+\frac{x+12}{1993}+1\)
\(\Leftrightarrow\frac{x+2005}{1999}+\frac{x+2005}{1997}=\frac{x+2005}{1995}+\frac{x+2005}{1993}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2005=0\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\ne0\right)\)
<=> 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, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(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
ok
\(\frac{x-30}{10}+\frac{x-28}{9}+\frac{x-26}{8}=-6\)
<=> \(\frac{36.\left(x-30\right)}{360}+\frac{40\left(x-28\right)}{360}+\frac{45\left(x-26\right)}{360}=\frac{-2160}{360}\)
=> \(36x-1080+40x-1120+45x-1170=-2160\)
\(< =>36x+40x+45x=-2160+1080+1120+1170\)
<=> \(121x=1210\)
<=> x = 10