`125,75-12,5xx x=100,25`

`12,5xx x=125,75-100,25`

`12,5xx x=25,5`

`x=25,5:12,5`

`x=2,04`

\(3,75+x\times0,5=18,25\\ x\times0,5=18,25-3,75\\ x\times0,5=14,5\\ x=14,5:0,5\\ x=29\)

.-. um 

từ thg 8 năm ngoái .-.

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}=\dfrac{x+3}{97}+\dfrac{x+4}{96}\\ \Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1=\dfrac{x+3}{97}+1+\dfrac{x+4}{96}+1\\ \Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}=\dfrac{x+100}{97}+\dfrac{x+100}{96}\\ \Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{96}\right)\\ \Leftrightarrow x+100=0\\ \Leftrightarrow x=-100\)

`a)` Ta có :

`BA=BC` \((gt)\)

`=>B in` đg trung trực `AC`

`DA=DC`\((gt)\)

`=>D in` đg trung trực `AC`

`=>BD` là đg trung trực `AC`

`b)` Xét \(\Delta BAD\) và \(\Delta BCD\) có :

`BA=BC`\((gt)\)

`DA=DC`\((gt)\)

`BD` chung

Do đó : \(\Delta BAD=\Delta BCD\left(c.c.c\right)\\ \Rightarrow\widehat{BAD}=\widehat{BCD}\)

\(\widehat{BAD}+\widehat{BCD}+\widehat{ABC}+\widehat{ADC}=360^0\\ \widehat{BAD}+\widehat{BCD}=360^0-\left(\widehat{ABC}+\widehat{ADC}\right)\\ 2\widehat{BAD}=360^0-\left(100^0+70^0\right)\\ 2\widehat{BAD}=190^0\\ \widehat{BAD}=\dfrac{190^0}{2}=95^0\\ \widehat{BAD}=\widehat{BCD}=95^0\)

\(a,\\ \dfrac{5}{11}\times\dfrac{7}{25}+\dfrac{5}{11}\times\dfrac{1}{5}\\ =\dfrac{5}{11}\times\left(\dfrac{7}{25}+\dfrac{1}{5}\right)\\ =\dfrac{5}{11}\times\dfrac{12}{25}\\ =\dfrac{1\times12}{11\times5}=\dfrac{12}{55}\\ b,\\ \dfrac{3}{7}\times\dfrac{25}{19}-\dfrac{1}{7}\times\dfrac{18}{19}\\ =\left(\dfrac{3}{7}-\dfrac{1}{7}\right)\times\left(\dfrac{25}{19}-\dfrac{18}{19}\right)\\ =\dfrac{2}{7}\times\dfrac{7}{19}\\ =\dfrac{2}{19}\)