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\(\left(2,8.x-32\right):\frac{2}{3}=90\)
\(2,8.x-32=90.\frac{2}{3}\)
\(2,8.x-32=60\)
\(2,8.x=60+32\)
\(2,8.x=92\)
\(x=92:2,8\)
\(x=\frac{230}{7}\)
a ) (2,8x - 32 ) : 2/3 = -90
<=> 2,8x -32 = -60
<=> 2,8x = -28
<=> x = -10
b ) 4/5 + 5/7 :x =1/6
<=> 5/7:x = -19/30
<=> x = -150/133
a)\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90\cdot\frac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy x = -10
b) \(\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)
\(\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}\)
\(\frac{5}{7}:x=\frac{-19}{30}\)
\(x=\frac{5}{7}:\frac{-19}{30}\)
\(x=\frac{-150}{133}\)
Vậy x = -150/133
=))
a)x+3/6=7/2
x =7/2-3/6
x =3
b)2,8+(3x-5/6)=2/2/5
14/5+(3x-5/6)=12/5
(3x-5/6)=12/5-14/5
3x+5/6=-2/5
3x =-2/5-5/6
3x =-37/30
x =-37/36:3
x =-37/90
A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
a) -0,6 . x - 7/3 = 5,4 (sửa dấu phẩy thành dấu nhân)
=> \(-\frac{3}{5}x-\frac{7}{3}=\frac{27}{5}\)
=> \(-\frac{3}{5}x=\frac{27}{5}+\frac{7}{3}=\frac{116}{15}\)
=> \(x=\frac{116}{15}:\left(-\frac{3}{5}\right)=\frac{116}{15}\cdot\left(-\frac{5}{3}\right)=-\frac{116}{9}\)
b) 2,8 : \(\left(\frac{1}{5}-3x\right)=1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=2,8:1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=\frac{14}{5}:\frac{7}{5}=\frac{14}{5}\cdot\frac{5}{7}=2\)
=> 3x = \(\frac{1}{5}-2=-\frac{9}{5}\)
=> \(x=\left(-\frac{9}{5}\right):3=\left(-\frac{9}{5}\right)\cdot\frac{1}{3}=-\frac{3}{5}\)
Nên sửa lại đề câu a đi nhá :))
\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)
\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)
\(\Leftrightarrow2,8x=-60+32=-28\)
\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)
d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)
Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.
mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha
(2,8.x-32):2/3=-90
=>2,8.x-32=90x2/3
=>2,8.x-32=60
=>2,8.x=60+32
=>2,8.x=92
=>x=92:2,8
=>x=230/7
( 2,8.x - 32 ) : \(\frac{2}{3}\) = - 90
= 2,8.x - 32 = 90. \(\frac{2}{3}\)
= 2,8.x - 32 = 60
= 2,8.x = 60 + 32
= 2,8.x = 92
x = 92 : 2,8
x = \(\frac{230}{7}\)
Vậy x bằng \(\frac{230}{7}\)