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Câu 2:
a: x-158=32
=>x=158+32
=>x=190
b: \(x\cdot24=264\)
=>\(x=\dfrac{264}{24}\)
=>x=11
c: \(6x+9=3^7:3^4\)
=>\(6x+9=3^3\)
=>6x+9=27
=>6x=18
=>x=18/6=3
Câu 1:
a: \(86\cdot19+14\cdot19\)
\(=19\left(86+14\right)\)
\(=19\cdot100=1900\)
b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)
\(=4\cdot25-104\cdot25\)
\(=25\left(4-104\right)=-100\cdot25=-2500\)
c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)
\(=7\cdot2\cdot8\cdot5\)
\(=56\cdot10=560\)
d: \(59-\left[59+\left(-76\right)\right]\)
\(=59-59+76\)
=76
\(\Rightarrow7x=5^{11-9}+3\)
\(\Rightarrow7x=5^2+3\)
\(\Rightarrow7x=28\)
\(\Rightarrow x=\frac{28}{7}\)
6x + x = 511 : 59 + 31
x . ( 6 + 1 ) = 52 + 3
x . 7 = 25 + 3
x . 7 = 28
x = 28 : 7
x = 4
Vậy x = 4
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
6x = 511 : 59 + 31 - x
<=> 6x = 52 + 3 - x
<=> 6x = 25 + 3 - x
<=> 6x + x = 28
<=> 7x = 28
=> x = 28 : 7
=> x = 4
Vậy x = 4
Thank ban