Bài 1: Tìm x
a,\(\left|3x-1\right|\)=5
b,\(\left|2x-3\right|\)-1=0
c,2\(\left|2x-3\right|\)=\(\dfrac{1}{2}\)
d,7-3\(\left|5-2x\right|\)=1
e,\(\left|x-\dfrac{2}{7}\right|\)=2+\(\dfrac{1}{3}\)
f,3(\(\dfrac{1}{2}\)-x)+\(\dfrac{1}{3}\)=\(\dfrac{7}{6}-x\)
g,2(\(\dfrac{3}{2}-x\))-\(\dfrac{1}{3}=7x-\dfrac{1}{4}\)
h,\(\dfrac{3}{2}-4\left(\dfrac{1}{4}-x\right)=\dfrac{2}{3}-7x\)
Bài 2:Tìm x
a,\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
BÀI:11;13;16(SGK TRANG 12,13)
CÁC BẠN GIÚP MK VỚI NHA MK CẦN GẤP
Tìm x dễ thì tự làm nha:
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)-\left(\dfrac{x+2}{2002}+1\right)-\left(\dfrac{x+1}{2003}\right)=0\)\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
