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Ta có
\(\frac{x-7}{6}+1+\frac{x-11}{10}+1=\frac{x-8}{7}+1+\frac{x-10}{9}+1\)
\(\frac{x-1}{6}+\frac{x-1}{10}-\frac{x-1}{7}-\frac{x-1}{9}=0\)
<=>\(\left(x-1\right)\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{7}-\frac{1}{11}\right)=0\)
=>x-1=0
<=>x=1
<=>3(x-3)=2(x-2)
<=>3x-9=2x-4
<=>3x-2x=-4+9
<=>x=5
tick nha
\(x\ne-\frac{7}{2};x\ne-3\)
\(\Rightarrow3\left(3x+9\right)=5\left(2x+7\right)\)
\(\Rightarrow9x+27=10x+35\)
\(\Rightarrow10x-9x=27-35\)
\(\Rightarrow x=-8\)
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)và \(5x+y-2z=28\)
\(\frac{1}{a+1}-\frac{1}{a+7}=\frac{a+7}{\left(a+1\right)\left(a+7\right)}-\frac{a+1}{\left(a+1\right)\left(a+7\right)}=\frac{6}{\left(a+1\right)\left(a+7\right)}\)
=>x+7=6
=>x=6-7
=>x=-1
vậy x=-1
\(\frac{1}{a+1}-\frac{1}{a+7}=\frac{\left(a+7\right)-\left(a+1\right)}{\left(a+1\right)\left(a+7\right)}=\frac{6}{\left(a+1\right)\left(a+7\right)}\)=> x + 7 = 6 => x = -1
\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\))
\(\Leftrightarrow x=-1\)
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
\(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\)
\(\Rightarrow\left(\frac{x-7}{6}+1\right)+\left(\frac{x-11}{10}+1\right)=\left(\frac{x-8}{7}+1\right)+\left(\frac{x-10}{9}+1\right)\)
\(\Rightarrow\frac{x-1}{6}+\frac{x-1}{10}-\frac{x-1}{7}-\frac{x-1}{9}=0\)
\(\left(x-1\right).\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)