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8 tháng 9 2017

a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)

\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)

\(=\frac{5-2^4.10}{2}\)

\(=5-8.10\)

\(=5-80\)

\(=-75\)

8 tháng 9 2017

a ) = -0,7499957912

b ) = -0,75

7 tháng 12 2018

\(a,\frac{4^5.2^{16}}{16^6}=\frac{2^{10}.2^{16}}{2^{24}}=\frac{2^{26}}{2^{24}}=2^2=4\)

\(b,\left(1-\frac{2}{5}\right)^2+|\frac{-3}{5}|+\frac{-7}{10}=\frac{9}{25}+\frac{3}{5}+\frac{-7}{10}=\frac{24}{25}-\frac{7}{10}=\frac{13}{50}\)

c, tt

23 tháng 7 2019

a) Vì\(x=99\Rightarrow x+1=100\)

Thay x+1=100 vào biểu thức A ta được :

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+9\)

\(=x+9\)

\(=99+9\)

\(=108\)

b) Tương tự

23 tháng 7 2019

\(A=x^5-100x^4+100x^3-100x^2+100x-9\)

\(\Rightarrow A=x^5-99x^4-x^4+99x^3+x^3-99x^2-x^2+99x+x-9\)

\(\Rightarrow A=x^4\left(x-99\right)-x^3\left(x-99\right)+x^2\left(x-99\right)+x\left(x-99\right)-9\)

\(\Rightarrow A=x^4\left(99-99\right)-x^3\left(99-99\right)+x^2\left(99-99\right)+x\left(99-99\right)-9\)

\(\Rightarrow A=x^4.0-x^3.0+x^2.0+x.0-9\)

\(\Rightarrow A=0-0+0+01-9=-9\)

22 tháng 2 2018

Ta có : 

\(A=\frac{\left(a+1\right)\left(a+2\right)\left(a+3\right).....\left(a+2003\right)\left(a+2004\right)}{\left(b+5\right)\left(b+6\right)\left(b+7\right).....\left(b+2006\right)\left(b+2007\right)}\)

\(\Leftrightarrow\)\(A=\frac{\left(0+1\right)\left(0+2\right)\left(0+3\right).....\left(0+2003\right)\left(0+2004\right)}{\left(-4+5\right)\left(-4+6\right)\left(-4+7\right).....\left(-4+2006\right)\left(-4+2007\right)}\)

\(\Leftrightarrow\)\(A=\frac{1.2.3.....2003.2004}{1.2.3.....2002.2003}\)

\(\Leftrightarrow\)\(A=\frac{1.2.3.....2003}{1.2.3.....2003}.2004\)

\(\Leftrightarrow\)\(A=2004\)
 

Vậy \(A=2004\)

a: \(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)

\(=\left(4\cdot32\right):\left(\frac{8}{16}\right)\)

\(=128:\frac12=128\cdot2=256\)

b: \(B=3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)

\(=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=\frac{3^8}{3^8}\cdot3^2=3^2=9\)

c: \(D=\left\lbrace\left(0,1\right)^2\right\rbrace^0+\left\lbrack\left(\frac17\right)^1\right\rbrack^2:\frac{1}{49}\cdot\left\lbrack\left(2^2\right)^3:2^5\right\rbrack\)

\(=1+\left(\frac17\right)^2\cdot49\cdot2^6:2^5\)

\(=1+49\cdot\frac{1}{49}\cdot2=1+2=3\)

d: \(C=\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\frac{17}{2}\right)^7:\left(\frac{17}{2}\right)^6\)

\(=\left(-0,5\right)^{5-3}-\left(\frac{17}{2}\right)\)

\(=\left(-0,5\right)^2-\frac{17}{2}=0,25-\frac{17}{2}=\frac14-\frac{34}{4}=-\frac{33}{4}\)

14 tháng 6 2017

\(A=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)

\(A=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)

\(A=\left(5-6-2\right)\)\(-\left(\dfrac{3}{4}+\dfrac{7}{4}-\dfrac{5}{4}\right)\)\(+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)

\(A=-3\)\(-\dfrac{5}{4}\)\(+\dfrac{-7}{5}\)

\(A=\dfrac{-113}{20}\)

14 tháng 6 2017

\(A=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)

\(A=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)

\(A=-3-\dfrac{5}{4}+\dfrac{-7}{5}\)

\(A=-\dfrac{113}{20}\)