giải toán nâng cao mình với:
tìm x biết :
a, 3\(\dfrac{1}{2}\)-\(\dfrac{1}{2}\)x=\(\dfrac{2}{3}\) ; b,\(\dfrac{1}{3}\)+\(\dfrac{2}{3}\):x=-7 ; c, \(\dfrac{1}{3}\)x+\(\dfrac{2}{5}\)(x-1)=0 ; d, x:\(\dfrac{3}{4}\)+\(\dfrac{1}{4}\)=\(\dfrac{-2}{3}\) ;
e, 2/\(\dfrac{1}{2}\)x-\(\dfrac{1}{3}\)/-\(\dfrac{3}{4}\)=\(\dfrac{1}{4}\) ; f, (2x-3)(6-2x)=0
cái dấu / là dấu giá trị tuyệt đối
giúp với !!!!!!!!! mình cần gấp chiều thứ năm phải có bài rồi
tìm x:
a)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
b) \(\left(x.6\dfrac{2}{7}+\dfrac{3}{7}\right).2\dfrac{1}{5}-\dfrac{3}{7}=-2\)
c) \(x.3\dfrac{1}{4}+\left(\dfrac{-7}{6}\right).x-1\dfrac{2}{3}=\dfrac{5}{12}\)
d) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
tìm x biết
a) \(\dfrac{2x-3}{3}+-\dfrac{3}{2}=\dfrac{5-2x}{6}-\dfrac{1}{2}\)
b) \(\left(3x-\dfrac{1}{2}\right).\left(3-\dfrac{1}{2}x\right).\left(1+\dfrac{3}{4}:x\right)=0\)
c) \(\left(3-\dfrac{9}{10}-|x+2|\right):\left(\dfrac{19}{20}-1-\dfrac{2}{3}\right)+\dfrac{4}{5}=1\)
Tìm x biết :
1/\(\dfrac{4}{3}\) + x = \(\dfrac{2}{15}\) 2/\(\dfrac{1}{3}+\dfrac{1}{2}:\) x = -0,25 3/(\(\dfrac{2x}{5}+2\) ):(-4) = -1\(\dfrac{1}{2}\) 4/ x.( x + 1) = 0 5/3\(\dfrac{1}{2}-\dfrac{1}{2}x\) = \(\dfrac{2}{3}\)
6/1\(\dfrac{1}{2}+x=\dfrac{3}{2}-7\) 7/\(2x-16=40+x\) 8/\(x:\dfrac{5}{3}+\dfrac{1}{3}=\dfrac{-2}{5}\) 9/\(\dfrac{x}{4}=\dfrac{9}{x}\)
BT3: Tìm x, biết
15) \(2\text{|}\dfrac{1}{2}x-\dfrac{1}{3}\text{|}-\dfrac{3}{2}=\dfrac{1}{4}\)
16) \(\dfrac{3}{4}-2.\text{|}2x-\dfrac{2}{3}\text{|}=2\)
(Lưu ý: Dấu '' | '' là dấu giá trị tuyệt đối)
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b,\(60\%.x+\dfrac{2}{3}.x=\dfrac{1}{3}.6\dfrac{1}{3}\)
c, \(\left(\dfrac{9}{2}.x-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
d,\(\left(x+\dfrac{5}{6}\right).2\dfrac{2}{5}-1\dfrac{1}{4}=35\%\)
tìm x biết
\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(\left(0.6-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(\left(3x-1\right).\left(-\dfrac{1}{2}x+5\right)=0\)
\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
Tìm số x , biết
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=-\dfrac{53}{10}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
d) \(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(60\%x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\left(2x+\dfrac{3}{2}\right)^2-\dfrac{9}{25}=0\)
\(60\%x+0,4x+x:=2\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(-2x-\dfrac{-3}{5}:\left(-0,5\right)^2=-1\dfrac{1}{4}\)
Tìm x
a. \(\dfrac{-15}{12}\) x+ \(\dfrac{3}{7}\) = \(\dfrac{6}{5}\) x − \(\dfrac{1}{2}\)
b, / \(\dfrac{1}{2}-\dfrac{1}{3}\) + x/ = \(\dfrac{-1}{4}\) −/ y /
c, / \(\dfrac{9}{25}\)+y/ + / x - y / = 0
d, / /x + 5 / - 4 / = 3 ( câu này có 2 dấu giá trị tuyệt đối nhé chứ ko pải viết thừa đâu nhé )
Lưu ý : dấu / là giá trị tuyệt đối nhé