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a) Ta có: \(\left(2^2\right)^3\cdot4^5\)
\(=2^6\cdot2^{10}\)
\(=2^{16}=65536\)
b) Ta có: \(\left[\left(-4\right)^2\right]^2\cdot6\)
\(=16^2\cdot6\)
\(=256\cdot6=1536\)
c) Ta có: \(\frac{16}{25}\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^2\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^5\)
\(=\frac{1024}{3125}\)
d) Ta có: \(\left(\frac{121}{64}\right)^2\cdot\left(-\frac{64}{11}\right)^2\)
\(=\frac{121^2}{64^2}\cdot\frac{64^2}{11^2}\)
\(=11^2=121\)
e) Ta có: \(\left[\left(-3\right)^3\right]^3\cdot271:125\)
\(=\left(-27\right)^3\cdot\frac{271}{125}\)
\(=\frac{-5334093}{125}\)
(4 . 9)15 < (2 . 3)x < (18 . 2)6
3615 < 6x < 366
(62)15 < 6x < (62)6
630 < 6x < 612
=> 30 < x < 12
=> x ko tồn tại
Ta có quy luật:
1+1=12 = 1
2+2=22 = 4
3+3 = 32 = 9
..................
.......................
9 + 9 = 92 = 81
\(5-\left|3x-1\right|=3\)
\(\left|3x-1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
\(\left|x+\frac{3}{4}\right|-5=-2\)
\(\left|x+\frac{3}{4}\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{15}{4}\end{cases}}\)
\(\left(1-2x\right)^2=9\)
\(\left(1-2x\right)^2=3^2\)
\(\Rightarrow1-2x=3\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
vậy \(x=-1\)
\(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
vậy \(x=-9\)
\(\left(2x+1\right)^2=\frac{4}{9}\)
\(\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow2x+1=\frac{2}{3}\)
\(\Rightarrow2x=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{6}\)
vậy \(x=-\frac{1}{6}\)
\(A=\left(8+2\cdot3-7\cdot\dfrac{13}{10}+3\cdot\dfrac{5}{4}\right):\left(\dfrac{5\sqrt{6}}{3}\right)^2\\ A=\left(14-\dfrac{91}{10}+\dfrac{15}{4}\right):\dfrac{50}{3}\\ A=\dfrac{173}{20}\cdot\dfrac{3}{50}=\dfrac{519}{1000}\)
S = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/2019^2
S = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/2019^2)
S = 2018 - (1/4 + 1/9 + 1/16 + ... + 1/2019^2)
đặt A = 1/4 + 1/9 + 1/16 + ... + 1/2019^2
có : 1/4 = 1/2*2 < 1/1*2
1/9 = 1/3*3 < 1/2*3
...
1/2019^2 < 1/2018*2019
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + /12018*2019
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4+ ... + 1/2018 - 1/2019
=> A < 1 - 1/2019
=> A < 2018/2019
=> A không phải số nguyên
S = 2018 - A
=> S không phải 1 số nguyên
