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\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)

\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{3}{2}\)

\(=1\)

\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)

\(\Rightarrow\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)

a) \(\left(-\frac{1}{4}\right)^0=1\)

b) \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)

c) \(\left(\frac{4}{5}\right)^{-2}=\frac{25}{16}\)

d) \(\left(0,5\right)^{-3}=8\)

e) \(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)

a, \(\left(\frac{-1}{4}\right)^0\) = 1

Bất kỳ số nguyên nào nếu có mũ bằng 0 đều bằng 1

b, \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)

Câu a:

1 + {-2 - [-3 + (-4 + |x|)]} = 1 - 2 + [(-3 - 4)]

1 + {-2 -[-3 - 4 + |x|]} = 1 - 2 + [-7]

1 + {-2 - [- 7 + |x|]} = 1 - 2 - 7

1 + {- 2 + 7 - |x|} = - 1 - 7

1 + {5 - |x|} = - 8

5 - |x| = - 8 - 1

5 - |x| = - 9

|x| = 5 + 9

|x| = 14

x = - 14 hoặc x = 14

Vậy x ∈ {-14; 14}

Câu b:

34 + (9 - 21) = 3417 - (x + 3417)

34 + (-12) = 3417 - x - 3417

34 - 12 = 3417 - 3417 - x

22 = - x

x = 22 : (-1)

x = - 22

Vậy x = - 22

1. 

a.\(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

b. \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)

c. \(\left(\frac{-3}{5}\right)^5=\frac{-243}{3125}\)

d. \(\left(\frac{-1}{5}\right)^2=\frac{1}{25}\)

e. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)

Trả lời:

Bài 1: 

a, \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)

b, \(\left(\frac{1}{2}\right)^3=\frac{1^3}{2^3}=\frac{1}{8}\)

c, \(\left(\frac{-3}{5}\right)^2=\frac{\left(-3\right)^2}{5^2}=\frac{9}{25}\)

d, \(\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)

e, \(\left(\frac{-1}{6}\right)^3=\frac{\left(-1\right)^3}{6^3}=\frac{-1}{216}\)

Bài 2:

a, \(\left(\frac{3}{2}\right)^2.\left(\frac{4}{3}\right)^2=\frac{9}{4}.\frac{16}{9}=4\)

b, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

c, \(\left(-\frac{1}{2}\right)^2.\left(\frac{2}{5}\right)^2=\frac{1}{4}.\frac{4}{25}=\frac{1}{25}\)

d, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

e, \(\left(-5\right)^3.\frac{1}{5}=-125.\frac{1}{5}=-25\)

f, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)

Câu a:

18 - |x - 1| = 2^2

18 - |x - 1| = 4

|x - 1| = 18 - 4

|x - 1| = 14

x - 1 = -14 hoặc x - 1 = 14

Th1: x - 1 = -14

x = - 14 + 1

x = -13

Th2: x - 1 = 14

x = 14 + 1

x = 15

Vậy x ∈ {-13; 15}

Câu b:

x - {55 - [49 + (-28 - x)]} = 13 - {47 + [25 - (32 - x)]}

x - {55 - [49 - 28 - x]} = 13 - {47 + [25 - 32 + x]}

x - {55 - 49 + 28 + x} = 13 - {47 + 25 - 32 + x}

x - {6 + 28 + x} = 13 - {72 - 32 + x}

x - {34 + x} = 13 - {40 + x}

x - 34 - x = 13 - 40 - x

(x - x) + x = 13 - 40 + 34

0 + x = - 27 + 34

x = 7

Vậy x = 7