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\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\))
\(\Leftrightarrow x=-1\)
Vậy x=-1
mỗi phân số + 1 thì sẽ có tử chung là x + 1
chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)) =0
mà tổng các phân số kia khác 0 nên x+1 bằng 0
=> x=-1
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))
2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)
\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)
Vậy \(x=\frac{-92}{105}\)
b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)
Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)