125\(\times\)(-61)\(\times\)(-2)\(^3\)\(\times\)(-1)\(^{2n}\) (n\(\in\)N)
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a: A=120-61x(-8)
=120+488=608
b: =47(-136+36)-(-1)
=-4700+1=-4699
c: C=(-48-304+72)+36
=-352+72+36
=-280+36=-244
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}.....\dfrac{30}{62}.\dfrac{31}{64}=2x\)
\(\dfrac{1.2.3.4.5.....30.31}{\left(2.2\right)\left(2.3\right)\left(2.4\right)\left(2.5\right)\left(2.6\right).....\left(2.31\right)\left(2.32\right)}=2x\)
\(\dfrac{1\left(2.3.4.5....30.31\right)}{32\left(2.3.4.5.....31\right).2^{31}}=2x\)
\(\dfrac{1}{2^5.2^{31}}=2x\Rightarrow2x=\dfrac{1}{2^{36}}\Rightarrow x=\dfrac{1}{2^{36}}\div2=\dfrac{1}{2^{37}}\)
Vậy x = \(\dfrac{1}{2^{37}}\)
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\\ \dfrac{1}{2\cdot2}\cdot\dfrac{2}{2\cdot3}\cdot\dfrac{3}{2\cdot4}\cdot\dfrac{4}{2\cdot5}\cdot...\cdot\dfrac{30}{2\cdot31}\cdot\dfrac{31}{2\cdot32}=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{30}{31}\cdot\dfrac{31}{32}\right)=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot3\cdot4\cdot5\cdot...\cdot31\cdot32}\right)=2x\\ \dfrac{1}{2}\cdot\dfrac{1}{32}=2x\\ 2x=\dfrac{1}{64}\\ x=\dfrac{1}{64}:2\\ x=\dfrac{1}{128}\)