\(=\frac{\left(-1\right)^7}{3^7}.3^7+\left(\frac{1}{8}\right)^3.8^3+\left(\frac{1}{5}\right)^3.\left(2.5\right)^3\)

\(=\left(-1\right)^7+\frac{1^3}{8^3}.8^3+\frac{1^3}{5^3}.2^3.5^3\)

\(=-1+1+2^3=2^3=8\)

a ) \(\left(-\frac{40}{52}.0,32.\frac{17}{20}\right):\frac{64}{75}\)

= \(\left(-\frac{16}{65}.\frac{17}{20}\right):\frac{64}{75}\)

= \(\left(-\frac{68}{325}\right):\frac{64}{75}\)

= \(\frac{-51}{208}\)

b ) \(-\frac{10}{11}.\frac{8}{9}+\frac{7}{18}.\frac{10}{11}\)

= \(\frac{10}{11}.\left(-\frac{8}{9}+\frac{7}{18}\right)\)

= \(\frac{10}{11}.\left(-\frac{1}{2}\right)\)

= \(\frac{-5}{11}\)

c ) \(\frac{45^{10}.5^{20}}{75^{15}}\)

= \(\frac{5^{10}.3^{20}.5^{20}}{5^{30}.3^{15}}\)

= \(\frac{5^{30}.3^{20}}{5^{30}.3^{15}}\)

= 3 ⁵

= 243

d ) ( - 0,125 ) ³ . 80 ⁴

= -80000

Đây nhé bé

Câu1

Vì \(\mid x \mid \geq 0 \Rightarrow \mid x \mid + 1 \geq 1\).
Do đó \(\left(\right. \mid x \mid + 1 \left.\right)^{10} \geq 1^{10} = 1\).

Suy ra:

\(A = \left(\right. \mid x \mid + 1 \left.\right)^{10} + 2023 \geq 1 + 2023 = 2024.\)

Dấu “=” chỉ xảy ra khi \(\mid x \mid = 0 \Leftrightarrow x = 0\).

\(\Rightarrow\) Giá trị nhỏ nhất của \(A\) là \(\boxed{2024}\), đạt tại \(x = 0\).

Câu 2 ( câu này kiến thức nâng cao nhé em nên là khi em đọc lời giải sẽ có khó hiểu nhé )

Đặt \(n = 2022\). Khi đó:

\(A = \frac{n^{2022} + 1}{n^{2023} + 1} , B = \frac{n^{2021} + 1}{n^{2022} + 1} .\)

Xét tổng quát với \(a_{k} = \frac{n^{k} + 1}{n^{k + 1} + 1} , \left(\right. n > 1 \left.\right)\).

Ta gọi k là luỹ thừa của cơ số

\(a_{k} > a_{k - 1} \textrm{ }\textrm{ } \Longleftrightarrow \textrm{ }\textrm{ } \left(\right. n^{k} + 1 \left.\right)^{2} > \left(\right. n^{k + 1} + 1 \left.\right) \left(\right. n^{k - 1} + 1 \left.\right) .\)

Xét hiệu:

\(\left(\right.n^{k}+1\left.\right)^2-\left(\right.n^{k+1}+1\left.\right)\left(\right.n^{k-1}+1\left.\right)=-n^{k-1}\left(\right.n-1\left.\right)^2<0\)

Vậy \(a_{k} < a_{k - 1}\), tức dãy \(\left(\right. a_{k} \left.\right)\) giảm dần theo \(k\)

Do đó:

\(A = a_{2022} < a_{2021} = B .\)

\(\Rightarrow B>A\)

Câu3

Ta đổi : \(27 = 3^{3}\), \(9 = 3^{2}\), \(125 = 5^{3}\).

\(\frac{5^{16} \cdot \left(\right. 3^{3} \left.\right)^{7}}{\left(\right. 5^{3} \left.\right)^{5} \cdot \left(\right. 3^{2} \left.\right)^{11}} = \frac{5^{16} \cdot 3^{21}}{5^{15} \cdot 3^{22}} = 5^{16 - 15} \cdot 3^{21 - 22} = \frac{5}{3} .\)

Vậy kết quả bằng \(\frac{5}{3}\).

Câu 3:

\(\frac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)

\(=\frac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}=\frac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)

\(=\frac53\)

Câu 2:

\(2022A=\frac{2022^{2023}+2022}{2022^{2023}+1}=1+\frac{2021}{2022^{2023}+1}\)

\(2022B=\frac{2022^{2022}+2022}{2022^{2022}+1}=1+\frac{2021}{2022^{2022}+1}\)

Ta có: \(2022^{2023}+1>2022^{2022}+1\)

=>\(\frac{2021}{2022^{2023}+1}<\frac{2021}{2022^{2022}+1}\)

=>\(\frac{2021}{2022^{2023}+1}+1<\frac{2021}{2022^{2022}+1}+1\)

=>2022A<2022B

=>A<B

Câu 1:

\(\left|x\right|\ge0\forall x\)

=>\(\left|x\right|+1\ge1\forall x\)

=>\(\left(\left|x\right|+1\right)^{10}\ge1^{10}=1\forall x\)

=>\(\left(\left|x\right|+1\right)^{10}+2023\ge1+2023=2024\forall x\)

Dấu '=' xảy ra khi x=0

Câu a:

(0,25)\(^3\).32

= (0,25)\(^3\).4\(^2\).2

= (0,25.4)\(^2\).(0,25.2)

= 1\(^2\).0,5

= 0,5

Câu b:

(-0,125)\(^3\).(80)\(^4\)

= (-0,125.80)\(^3\) .(80)

= -10\(^3\).80

= -80000

x:7/10=-4/5-3/50

x:7/10=-43/50

x=43/50.7/10

x=-301/500

Ta có: \(\frac{-4}{5}-x:\frac{7}{10}=\frac{3}{50}\)

\(\Rightarrow\)\(x:\frac{7}{10}=\frac{-43}{50}\)

\(\Rightarrow\)\(x=\frac{-301}{500}\)

Vậy\(x=\frac{-301}{500}\)

trả lời:

\(\left(\frac{9}{14}\right)^2.\left(-\frac{7}{3}\right)^3=\frac{3^2}{2^2.7^2}.\frac{-7^3}{3^3}\)

=\(\frac{3}{2^2}.\frac{-7}{3}=-\frac{7}{4}\)

\(\left(\frac{9}{14}\right)^2\cdot\left(-\frac{7}{3}\right)^3=\left(\frac{3\cdot3}{7\cdot2}\cdot\frac{-7}{3}\right)^2\cdot\frac{-7}{3}=\frac{9}{4}\cdot-\frac{7}{3}=\frac{-21}{4}\)