Bài 1 Tìm x ,y biết
a, 2| 2x -3 | =1
b,7,5 -3 | 5 -2x | = -4,5
c, | 3x-4 | + | 3y +5 |=0
Bài 2 Tìm x biết
a, \(\left|\frac{5}{3}x\right|\) =\(\left|-\frac{1}{6}\right|\)
b, \(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|-\frac{3}{4}\right|\)
c, \(\left|x+\frac{3}{5}\right|-\left|x-\frac{7}{3}\right|=0\)
Bài 3 Tìm x biết
a,| x |+| x+2 |=0
b,\(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\)
Bài 1 : Tính
1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)
2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)
3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)
4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)
5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)
6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)
Bài 2 : Tìm x, biết :
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
4) \(\left|x+\frac{2}{3}\right|=0\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
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E cần gấp lắm ạ, cảm ơn.
Bài 1 : Tính
1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)
2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)
3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)
4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)
5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)
6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)
Bài 2 : Tìm x, biết :
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
4) \(\left|x+\frac{2}{3}\right|=0\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
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E đang cần gấp, cảm ơn :)
Bài 2 : Tìm x, biết :
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
4) \(\left|x+\frac{2}{3}\right|=0\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Bài 1:
1. Tính: \(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
2. Tìm và tính tổng các số nguyên x thỏa mãn: \(\frac{21}{5}\left|x\right|< 2019\)
3. Tìm x, biết: \(\frac{2^{24}\left(x-3\right)}{\left(3\frac{5}{7}-1,4\right)\left(6\cdot2^{24}-4^{13}\right)}=\left(\frac{5}{3}\right)^2\)
Bài 1: Tìm x a) \(\frac{1}{7}\cdot\left|x\right|-\frac{4}{5}=\frac{1}{5}\) b) \(\left(\frac{3}{4}-x\right)^3=-27\) c) \(\frac{x-1}{-2}=\frac{-8}{x-1}\) d) \(\left|2x-3\right|+4\cdot\left(-5\right)^2=101\)
Bài 1: Tìm x
1)\(2\left|\frac{1}{2}.x-\frac{3}{8}\right|-\frac{3}{2}=\frac{1}{4}\)
2) -5.\(\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}.x-\frac{5}{6}\)
3) 3.\(\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
4) \(\frac{3}{4}-2.\left|2.x-0,125\right|=2\)
5) \(2.\left|\frac{1}{2}.x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
Cần gấp giúp mình với ai trả lời mình tick cho.
\(\frac{12}{16}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\) \((-0,6x-\frac{1}{2}).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\frac{-2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
4,Tìm x biết
a,\(\frac{-2}{3}\cdot x+\frac{1}{5}=\frac{3}{10}\)
b,\(\frac{2}{3}\cdot x-\frac{3}{2}\cdot x=\frac{5}{12}\)
c,\(\left(4,5-2x\right)\cdot\left(-1\frac{4}{7}\right)=\frac{11}{14}\)
d,\(\frac{1}{4}+\frac{1}{3}:3x=-5\)
e,\(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
g,\(|4x-1|=\left(-3^2\right)\)
h,\(|x+70|=2\frac{1}{5}\)
i,\(\left(x-1^3\right)=125\)
k,\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)
Giúp mình nhé!!!