\(\dfrac{-1}{\dfrac{4}{0,5-x}}=\dfrac{x-0,5}{4}\)
35,5 x \(\dfrac{-3}{4}\)- \(\dfrac{1}{4}\)x 35,5+0,5: \(\dfrac{-5}{6}\)
helpppppppppppppp
=35,5(-3/4-1/4)+1/2*(-6/5)
=-35,5-0,6
=-36,1
Tìm x, biết:
a) \(\left(\dfrac{3}{4}.x+2\dfrac{1}{2}\right).\dfrac{-2}{3}=\dfrac{1}{8}\)
b) \(\dfrac{1}{3}.x-0,5.x=0,75\)
a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)
=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)
=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)
=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)
b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)
=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)
=>\(x\cdot\dfrac{-1}{6}=0,75\)
=>\(x=-0,75\cdot6=-4,5\)
Tìm x, biết:
a) x+\(\dfrac{1}{6}\)=\(\dfrac{-3}{8}\) b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
c) \(\dfrac{1}{2}x\)+\(\dfrac{1}{8}x=\dfrac{3}{4}\) d) 75%-\(1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
Tìm x
\(\dfrac{1}{5}x:\dfrac{-4}{3}\left(0,5\right):\dfrac{-3}{4}\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
Tính bằng cách thuận tiện nhất :
a. 69,78 x \(\dfrac{3}{4}\) + 69,78 x \(\dfrac{1}{4}\) + 69,78 x 99
b. 34,1 x 8 + 34,1 x 0,5
\(a,=69,78\times\left(\dfrac{3}{4}+\dfrac{1}{4}+99\right)\\ =69,78\times\left(\dfrac{4}{4}+99\right)\\ =69,78\times\left(1+99\right)=69,78\times100=6978\\ b,34,1\times8+34,1\times0,5\\ =34,1\times\left(8+0,5\right)\\ =34,1\times8,5\\ =289,85\)
a: =69,78(3/4+1/4+99)
=69,78*100
=6978
b: =34,1(8+0,5)=34,1*8,5=289,85
a.\(69,78\times\dfrac{3}{4}+69,78\times\dfrac{1}{4}+69,78\times99=69,78\times\left(\dfrac{3}{4}+\dfrac{1}{4}+99\right)=69,78\times100=6978\)
b. \(34,1\times8+34,1\times0,5=34,1\times\left(8+0,5\right)=34,1\times8,5=289,85\)
1. tìm x
a, 2x - \(\dfrac{1}{2}\) : \(\dfrac{3}{4}\) = \(\dfrac{12}{9}\)
b, x\(^2\) - 2 = 0
2. Cho
A = \(\dfrac{1,11+0,19-13.2}{2,06+0,54}\) - \(\left(\dfrac{1}{2}+\dfrac{1}{4}\right)\) : 2
B = \(\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right)\) : \(2\dfrac{23}{26}\)
a, Rút gọn A, B
b, Tìm x ϵ Z để A < x < B
a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)
Bài 1:
a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)
=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)
=>\(2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
=>x=2/2=1
b: \(x^2-2=0\)
=>\(x^2=2\)
=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Bài 2:
a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)
\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)
\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)
\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)\)
\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)
b: A<x<B
=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)
mà \(x\in Z\)
nên \(x\in\left\{-9;-8;...;0;1\right\}\)
72x+72x+3=344 (5-x)(9x2-4)=0 \(\left|2-2x\right|\)-3,75=(-0,5)2
\(\sqrt{x-1}\)+\(\dfrac{2}{3}\)=1 (\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)x)2=2\(\dfrac{1}{4}\) giúp mình với!!!! cảm ơn nhìu=))❤
(5 - \(x\))(9\(x^2\) - 4) =0
\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}
72\(x\) + 72\(x\) + 3 = 344
72\(x\) \(\times\) ( 1 + 73) = 344
72\(x\) \(\times\) (1 + 343) = 344
72\(x\) \(\times\) 344 = 344
72\(x\) = 344 : 344
72\(x\) = 1
72\(x\) = 70
\(2x\) = 0
\(x\) = 0
Kết luận: \(x\) = 0
|2 - 2\(x\)| - 3,75 = (-0,5)2
|2 - 2\(x\)| - 3,75 = 0,25
|2- 2\(x\)| =0,25 + 3,75
|2 - 2\(x\)| = 4
\(\left[{}\begin{matrix}2-2x=-4\left(x>1\right)\\2-2x=4\left(x< 1\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=6\left(x\ge1\right)\\2x=-2\left(x\le1\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Kết luận: \(x\) \(\in\) { -1; 3}
2x-0,5=x+\(\dfrac{1}{4}\)
2x-0,5=x+\(\dfrac{1}{4}\)
<=> 2x-x=\(\dfrac{1}{4}+0,5\)
<=> x= 0,75