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a) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
b) \(109.5^2-3^2.25\)
\(=109.25-9.25\)
\(=25\left(109-9\right)\)
\(=25.100\)
\(=2500\)
c) \(\left[5^2.6-20.\left(37-2^5\right)\right]:10-20\)
\(=\left[5^2.6-20.\left(37-32\right)\right]:10-20\)
\(=\left(5^2.6-20.5\right):10-20\)
\(=\left(25.6-20.5\right):10-20\)
\(=\left(150-100\right):10-20\)
\(=50:10-20\)
\(=5-20\)
\(=-15\)
100:{250:[450-(4.53-32.25)]}
=100:{250:[450-(4.125-9.25)]}
=100;{250:[450-(500-225)]}
=100:{250:[450-275]
=100:{250:175}
=100:10/7
=70
a) \(100:\left\{250:\left[450-\left(4.5^3-25.4\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4.125-25.4\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
b) \(4\left(18-15\right)-\left(5-3\right).3^2\)
\(=4.3-2.3^2\)
\(=4.3-2.9\)
\(=12-18\)
\(=-6\)
100:{250:[450-(4.53 -25.4)]}
=100:{250:[450-(4.125-25.4)]}
=100:{250:[450-(500-100)]}
=100:{250:[450-400]}
=100:{250:50}
=100:5
=20
b)4.(18-15)-(5-3).32
=4.(18-15)-(5-3).9
=4.3-2.9
=12-18
=(-6)
=4.
\(\left(3n\right)^{100}\\ =3^{100}.n^{100}\\ =\left(3^4\right)^{25}.n^{100}\\ =81^{25}.n^{100}⋮81\)
Vậy \(\left(3n\right)^{100}⋮81\)
Chúc em học tốt!
Giải:
Có:
\(S=\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)+4^3\)
Ta nhân thấy rằng trong tích \(\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)\) có một thừa số bằng 0, đó là thừa số \(2018-2018\)
Mà trong một tích, nếu có một thừa số bằng 0 thì tích đó bằng 0
\(\Leftrightarrow\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)=0\)
\(\Leftrightarrow S=\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)+4^3=0+4^3=4^3=64\)
Vậy \(S=64\)
Chúc bạn học tốt!
S= (2018-1)(2018-2) .... (2018-2017) . 0 +43
=> S= 0 + 43 (Trong 1 tích có 1 thừa số bằng 0 thì tích đó bằng 0);
=>S= 4.4.4=64;
Vậy S=64
Ta có: ( x + 2)( x - 5) = -12
=> \(x+2\inƯ\left(-12\right);x-5\inƯ\left(-12\right)\)
mà Ư (-12) = \(\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+2\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\x-5\in\left\{"....."\right\}\end{matrix}\right.\)
Xét các t/h:
a, Ta có: \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^4}\right)^7=\left(\dfrac{1}{3}\right)^{28}=\dfrac{1}{3^{28}}\)
\(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3^5}\right)^6=\left(\dfrac{1}{3}\right)^{30}=\dfrac{1}{3^{30}}\)
Vì \(\dfrac{1}{3^{28}}>\dfrac{!}{3^{30}}\Rightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\Rightarrow\) \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
b, Ta có: \(\left(\dfrac{3}{8}\right)^5=\dfrac{3^5}{\left(2^3\right)^5}=\dfrac{243}{2^{15}}>\dfrac{243}{3^{15}}>\dfrac{125}{3^{15}}=\dfrac{5^3}{\left(3^5\right)^3}=\left(\dfrac{5}{243}\right)^3\)
\(\Rightarrow\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)