Bài 4 trang 76 SGK Đại số và Giải tích 12 Nâng cao

Thực hiện phép tính:

a) ${81}^{-0,75}+{\left(\frac{1}{125}\right)}^{\frac{-1}{3}}-{\left(\frac{1}{32}\right)}^{\frac{-3}{5}};$ 

b) $0,{001}^{\frac{-1}{3}}-{\left(-2\right)}^{-2}{.64}^{\frac{2}{3}}-8^{-1\frac{1}{3}}$ $+{\left(9^0\right)}^2;$

c) ${27}^{\frac{2}{3}}+{\left(\frac{1}{16}\right)}^{-0,75}-{25}^{0,5}$ 

d) ${\left(-0,5\right)}^{-4}-{625}^{0,25}-{\left(2\frac{1}{4}\right)}^{-1\frac{1}{2}}$ $+19{\left(-3\right)}^{-3}$

LG a

${81}^{-0,75}+{\left(\frac{1}{125}\right)}^{\frac{-1}{3}}-{\left(\frac{1}{32}\right)}^{\frac{-3}{5}};$ 

Lời giải chi tiết:

${81}^{-0,75}+{\left(\frac{1}{125}\right)}^{\frac{-1}{3}}-{\left(\frac{1}{32}\right)}^{\frac{-3}{5}}$

$={\left(3^4\right)}^{\frac{-3}{4}}+{\left({\left(\frac{1}{5}\right)}^3\right)}^{\frac{-1}{3}}-{\left({\left(\frac{1}{2}\right)}^5\right)}^{\frac{-3}{5}}$

$={\left(3\right)}^{-3}+{\left(\frac{1}{5}\right)}^{-1}-{\left(\frac{1}{2}\right)}^{-3}$

$=\frac{1}{27}+5-8=\frac{1}{27}-3=-\frac{80}{27}$

Cách khác:

LG b

$0,{001}^{\frac{-1}{3}}-{\left(-2\right)}^{-2}{.64}^{\frac{2}{3}}-8^{-1\frac{1}{3}}$ $+{\left(9^0\right)}^2;$

Lời giải chi tiết:

$0,{001}^{\frac{-1}{3}}-{\left(-2\right)}^{-2}{.64}^{\frac{2}{3}}-8^{-1\frac{1}{3}}+{\left(9^0\right)}^2$

$={\left({10}^{-3}\right)}^{-\frac{1}{3}}-2^{-2}.{\left(2^6\right)}^{\frac{2}{3}}-{\left(2^3\right)}^{-\frac{4}{3}}+1$

$=10-2^2-2^{-4}+1=7-\frac{1}{16}=\frac{111}{16}$

Cách khác:

LG c

${27}^{\frac{2}{3}}+{\left(\frac{1}{16}\right)}^{-0,75}-{25}^{0,5}$

Lời giải chi tiết:

${27}^{\frac{2}{3}}+{\left(\frac{1}{16}\right)}^{-0,75}-{25}^{0,5}$

$={\left(3^3\right)}^{\frac{2}{3}}+{\left(2^{-4}\right)}^{-\frac{3}{4}}-{\left(5^2\right)}^{\frac{1}{2}}$

$=3^2+2^3-5=12$

Cách khác:

LG d

${\left(-0,5\right)}^{-4}-{625}^{0,25}-{\left(2\frac{1}{4}\right)}^{-1\frac{1}{2}}$ $+19{\left(-3\right)}^{-3}$

Lời giải chi tiết:

${\left(-0,5\right)}^{-4}-{625}^{0,25}-{\left(2\frac{1}{4}\right)}^{-1\frac{1}{2}}$ $+19{\left(-3\right)}^{-3}$

$={\left({\left(-2\right)}^{-1}\right)}^{-4}-{\left(5^4\right)}^{\frac{1}{4}}-{\left({\left(\frac{3}{2}\right)}^2\right)}^{-\frac{3}{2}}$ $+\frac{19}{-27}$

$=2^4-5-{\left(\frac{3}{2}\right)}^{-3}-\frac{19}{27}$

$=11-\frac{8}{27}-\frac{19}{27}=10.$

Cách khác:

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