Hình như ở trang 8 mới đúng bn nhỉ ?
11. Tìm số hữu tỉ x, biết rằng :
a) (3/2).x+4=-12
b) (3/4)+(1/4) : x = -3
c) |3x-5|=4
d) (x+1)/10 + (x+1)/11 + (x+1)/12 = (x+1)/13 + (x+1)/14
e)* (x+4)/2000 + (x+3)/2001 = (x+2)/2002 + (x+1)/2003
K nha
bạn cho mk biết đề bai 13 và 17 được ko
a/ \(\frac{2}{3}x+4=-12\)
=> \(\frac{2}{3}x=-16\)
=> \(x=-16.\frac{3}{2}=-24\)
b/ \(\frac{3}{4}+\frac{\frac{1}{4}}{x}=-3\)
=> \(\frac{\frac{1}{4}}{x}=-3-\frac{3}{4}\)
=> \(\frac{\frac{1}{4}}{x}=\frac{-12-3}{4}\)
=> \(\frac{\frac{1}{4}}{x}=\frac{-15}{4}\)
=> \(-15x=4.\frac{1}{4}\)
=> \(-15x=1\)
=> \(x=\frac{-1}{15}\)
c/ \(\left|3x-5\right|=4\)
=> \(\orbr{\begin{cases}3x-5=4\\3x-5=-4\end{cases}}\)=> \(\orbr{\begin{cases}3x=9\\3x=1\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{3}\end{cases}}\)
d/ \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
=> \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
=> \(\frac{1}{10}\left(x+1\right)+\frac{1}{11}\left(x+1\right)+\frac{1}{12}\left(x+1\right)-\frac{1}{13}\left(x+1\right)-\frac{1}{14}\left(x+1\right)=0\)
=> \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\left(x+1\right)=0\)
=> x + 1 = 0
=> x = -1
e) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
=> \(\frac{x+4}{2000}+\frac{x+3}{2001}-\frac{x+2}{2002}-\frac{x+1}{2003}=0\)
=> \(\frac{1}{2000}\left(x+4\right)+\frac{1}{2001}\left(x+3\right)-\frac{1}{2002}\left(x+2\right)-\frac{1}{2003}\left(x+1\right)=0\)
=> \(\frac{1}{2000}\left(x+1+3\right)+\frac{1}{2001}\left(x+1+2\right)-\frac{1}{2002}\left(x+1+1\right)-\frac{1}{2003}\left(x+1\right)=0\)
=> \(\frac{1}{2000}\left(x+1\right)+3+\frac{1}{2001}\left(x+1\right)+2-\frac{1}{2002}\left(x+1\right)+1-\frac{1}{2003}\left(x+1\right)=0\)
=> \(\left(\frac{1}{2000}+3+\frac{1}{2001}+2-\frac{1}{2002}+1-\frac{1}{2003}\right)\left(x+1\right)=0\)
=> x + 1 = 0
=> x = -1
