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(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 \(\frac{230}{7}\)
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/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
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)=> 4.(90:x-8)=32-24=8
=>90:x-8=8:4=2
=>90:x=2+8=10
=>x=90:10=9
b)=> 3x+15+2x+20=100
=> 5x+35=100
=>5x=100-35=65
=>x=13
nha . Chúc bạn học tốt
a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(2^{3x}:2^{x+3}=32\Rightarrow2^{3x-x-3}=2^5\Rightarrow x^{2x-3}=2^5\Rightarrow2x-3=5=>2x=8=>x=4\)
\(2^{5x}:2^{4x}=32\Rightarrow2^{5x-4x}=2^5=>2^x=2^5=>x=5\)
\(\left(x-\frac{1}{2}\right)^3=8=2^3\Rightarrow x-\frac{1}{2}=2\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}\)
Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0
b) \(\left(x-140\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(x-140=21\)
\(x=161\)
vay \(x=161\)
c) \(x-6:2-\left(48-24\right):2:6-3=0\)
\(x-3-24:2:6-3=0\)
\(x-3-2-3=0\)
\(x-8=0\)
\(x=8\)
vay \(x=8\)
d) \(x+5.2-\left(32+16.3:6-15\right)=0\)
\(x+10-\left(32+8-15\right)=0\)
\(x+10-25=0\)
\(x-15=0\)
\(x=15\)
vay \(x=15\)
a) \(390-\left(x-8\right)=168:13\)
\(390-x+8=\frac{168}{13}\)
\(x+8=390-\frac{168}{13}\)
\(x+8=\frac{5070}{13}-\frac{168}{13}\)
\(x+8=\frac{4902}{13}\)
\(x=\frac{4902}{13}-8\)
\(x=\frac{4798}{13}\)
vay \(x=\frac{4798}{13}\)
\(a.2^x=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
\(b.\left(x-5\right)^3=8\Leftrightarrow\left(x-5\right)^3=2^3\Leftrightarrow x-5=2\Leftrightarrow x=7\)
\(c.5^{x+1}=25\Leftrightarrow5^{x+1}=5^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)
\(d.3\cdot\left(6x+3\right)=3^5\)
\(\Leftrightarrow6x+3=3^4\)
\(\Leftrightarrow6x=3^4-3=81-3=78\)
\(\Leftrightarrow x=78:613\)
1a/ \(\left(15-x\right)+\left(x-12\right)=7-\left(-5+x\right)\)
=> \(\left(15-x\right)+\left(x-12\right)+\left(-5+x\right)=7\)
=> \(15-x+x-12-5+x=7\)
=> \(\left(15-12-5\right)-\left(x+x+x\right)=7\)
=> \(\left(15-12-5\right)-7=3x\)
=> \(3x=-2-7\)
=> \(3x=-9\)
=> \(x=\frac{-9}{3}=-3\)
b/ \(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
=> \(x-57-42-23-x=13-47+25-32+x\)
=> \(x-x+x=13-47+25-32+57+42+23\)
=> \(x=\left(13+23\right)-\left(47+57\right)+\left(25+57\right)-\left(32+42\right)\)
=> \(x=36-104+82-74\)
=> \(x=-60\)
d/ \(\left(x-3\right)\left(2y+1\right)=7\)
Vì 7 là số nguyên tố nên ta có 2 trường hợp:
TH1: \(\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}}\)=> \(\hept{\begin{cases}x=4\\y=3\end{cases}}\).
TH2: \(\hept{\begin{cases}x-3=7\\2y+1=1\end{cases}}\)=> \(\hept{\begin{cases}x=10\\y=0\end{cases}}\).
Các cặp (x, y) thoả mãn điều kiện: \(\left(4;3\right),\left(10;0\right)\).
