1.

a) \(-\frac{8}{27}=-\left(\frac{2}{3}\right)^3\)

b) \(\frac{81}{625}=\left(\frac{3}{5}\right)^4\)

2.

a) 27.81=2187=3⁷

b) sai đề

a. \(25.5^3.\frac{1}{625}.5^2=5^2.5^3.\frac{1}{5^4}.5^2=\frac{5^7}{5^4}=5^3\)

b. \(4.32:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{2^4}=\frac{2^4}{2^4}=1\)

c. \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.3^2.\frac{1}{5^2}==\frac{5^2}{5^2}.3^7=3^7\)

d. \(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2=\frac{1}{7^3}.7^4=\frac{7^4}{7^3}=7\)

Bài 1 :

a) (2x + 1)³ = 125

=> (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

b) (x - 5)⁴ = (x - 5)⁶

Với hai mũ khác nhau , ta chỉ có thể tìm được giá trị biểu thức bằng 1 hoặc 0 (giá trị của chúng bằng nhau)

+) (x - 5)⁴ = (x - 5)⁶ = 0

=> (x - 5)⁴ = 0

=> (x - 5)⁴ = 0⁴

=> x - 5 = 0 => x = 0 + 5 = 5

+) (x - 5)⁴ = (x- 5)⁶ = 1

=> (x - 5)⁴ = 1

=> (x - 5)⁴ = 1⁴

=> x - 5 = 1

=> x = 1 + 5

=> x = 6

Bài 4 :

a³ . a⁹ = a^{3 + 9} = a¹²

(a⁵)⁷.(a⁶)⁴ .a¹² = a³⁵ . a²⁴ . a¹² = a^{35 + 24 + 12} = a⁷¹

4.5² - 2.3² = 4.25 - 2.9

= 100 - 18

= 82

mong cac ban giup, minh can gap lam,tuy minh trinh bay hoi xau nhung mong cac ban giup

3.viet cac tich, thuong sau duoi dang luy thua:

a) \(\dfrac{2^{10}}{8^3}\)

\(=\dfrac{2^{10}}{\left(2^3\right)^3}\)

\(=\dfrac{2^{10}}{2^9}\)

\(=2^1\)

giúp tui với : -8 . 25 . ( -2 ) . ( -25 ) .4 .125

\(\text{a)Để C đạt GTNN}\)

\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)

\(\Rightarrow C\ge-10\)

\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)

b)\(\text{Để D đạt GTLN}\)

=>(2x-3)²+5 đạt GTNN

Mà (2x-3)²\(\ge\)5

\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)

@Nguyễn Việt Lâm

https://hoc24.vn/id/2782086

Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)

Thật vậy, BĐT trên tương đương:

\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)

\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)

\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)

Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)

\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)

\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(a.\left(\frac{2}{5}\right)^5:\left(\frac{9}{25}\right)^5=\left(\frac{2\cdot25}{9\cdot5}\right)^5=\frac{10}{9}^5\)

\(b.25\cdot5^3\cdot\frac{1}{625}\cdot5^2=\frac{5^7}{5^4}=5^3\)

\(c.\frac{20^5\cdot5^{10}}{100^5}=\frac{2^{10}\cdot5^{15}}{2^{10}\cdot5^{10}}=5^5\)

\(d.\frac{1}{7}^2\cdot\frac{1}{7}\cdot49^2=\frac{7^4}{7^3}=7\)