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Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)
\(\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=-5+5=0\)
a)\(\frac{x}{108}=\frac{-7}{9}.\frac{5}{6}\)
\(\frac{x}{108}=\frac{-35}{54}\)
\(\frac{x}{108}=\frac{-70}{108}\)
\(x=-70\)
b)
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\left(\frac{x+1}{99}+\frac{99}{99}\right)+\left(\frac{x+3}{97}+\frac{97}{97}\right)+\left(\frac{x+5}{95}+\frac{95}{95}\right)=\left(\frac{x+2}{98}+\frac{98}{98}\right)+\left(\frac{x+4}{96}+\frac{96}{96}\right)+\left(\frac{\left(x+6\right)}{94}+\frac{94}{94}\right)\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{92}+\frac{x+100}{94}+\frac{x+100}{96}\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{92}-\frac{x+100}{94}-\frac{x+100}{96}=0\)
\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\right)=0\)
\(Mà\) \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\ne0\)
Nên x+ 100 = 0
x = 0 - 100 = -100
Vậy x= -100
cộng 1 vào mỗi tỉ số,ta được:
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)\(\Rightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+2+98}{98}+\frac{x+4+96}{96}+\frac{x+6+94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{98}+\frac{x+100}{96}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{98}-\frac{x+100}{96}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\ne0\)
=>x+100=0
=>x=-100
Vậy x=-100
\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)
=> \(\frac{2}{3}-x=\frac{1}{3}\)
=> \(x=\frac{1}{3}\)
\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)
=> \(\frac{10}{12}\div x=\frac{1}{6}\)
=> \(x=\frac{60}{12}=5\)
\(\frac{x-12}{4}=\frac{1}{2}\)
=> \(\left(x-12\right)\cdot2=4\cdot1\)
=> \(2x-24=4\)
=> \(2x=28\)
=> \(x=14\)
\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)
\(\frac{x}{15}=\frac{-1}{15}\)
=> \(x=-1\)
\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)
\(\frac{x}{182}=\frac{-15}{91}\)
=> \(91x=182\cdot\left(-15\right)\)
=> \(91x=-2730\)
=> \(x=-30\)
g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)
\(\Leftrightarrow x=-1\)
h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)
\(\Leftrightarrow x=-30\)
\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)
\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)
Vì 1/49+1/48+...+1/45 khác 0
Nên x+50=0
do đó x=-50
