a: ta có: \(x^3=125\)
nên x=5
b: Ta có: \(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)
\(\Leftrightarrow16x=90+30-40=80\)
hay x=5
c: Ta có: \(31-2x=1+16:2^3\)
\(\Leftrightarrow31-2x=1+2=3\)
hay x=14
a. x3 = 125
<=> x = \(\sqrt[3]{125}\)
<=> x = 25
b. 16x + 40 = 10 . 32 + 5(1 + 2 + 3)
<=> 16x + 40 = 10 . 9 + 5 + 10 + 15
<=> 16x + 40 = 90 + 5 + 10 + 15
<=> 16x = 90 + 5 + 10 + 15 - 40
<=> 16x = 80
<=> x = 5
c. 31 - 2x = 123 + 16 : 23
<=> 31 - 2x = 1 + 16 : 8
<=> 31 - 1 - 2 = 2x
<=> 28 = 2x
<=> x = 14
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