a: \(\left(2x-1\right)^4=81\)

=>\(\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b:Sửa đề: \(\left(x-1\right)^5=-32\)

=>\(\left(x-1\right)^5=\left(-2\right)^5\)

=>x-1=-2

=>x=-1

c: \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

=>\(\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

=>\(\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

=>\(\left(2x-1\right)^6\cdot\left(2x-1-1\right)\cdot\left(2x-1+1\right)=0\)

=>\(2x\left(2x-1\right)^6\cdot\left(2x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

a) 

\(\left(2x-1\right)^4=81\\ \Rightarrow\left(2x-1\right)^4=3^4\)

TH1: 2x - 1 = 3 => 2x = 4 => x = 2 

TH2: 2x - 1 = -3 => 2x = -3 + 1 = -2 => x = -1 

b)

\(\left(x-1\right)^5=-32\\ \Rightarrow\left(x-1\right)^5=\left(-2\right)^5\\ \Rightarrow x-1=-2\\ \Rightarrow x=-2+1\\ \Rightarrow x=-1\)

c) 

\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\\\Rightarrow \left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

TH1: 

\(\left(2x-1\right)^6=0\\ \Rightarrow2x-1=0\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\)

TH2: 

\(\left(2x-1\right)^2-1=0\\ \Rightarrow\left(2x-1\right)^2=1\\ \Rightarrow\left(2x-1\right)^2=1^2\)

+) 2x - 1 = 1 => 2x = 2 => x = 1

+) 2x - 1 = -1 => 2x = 0 => x = 0

1: \(\left(x+1\right)^2=x^2+2\cdot x\cdot1+1^2=x^2+2x+1\)

2: \(\left(4+x\right)^2=4^2+2\cdot4\cdot x+x^2=16+8x+x^2\)

5: \(\left(5x+1\right)^2=\left(5x\right)^2+2\cdot5x\cdot1+1^2=25x^2+10x+1\)

6: \(\left(2x+3\right)^2=\left(2x\right)^2+2\cdot2x\cdot3+3^2=4x^2+12x+9\)

9: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\)

10: \(\left(x+5y\right)^2=x^2+2\cdot x\cdot5y+\left(5y\right)^2=x^2+10xy+25y^2\)

13: \(\left(3x+5y\right)^2=\left(3x\right)^2+2\cdot3x\cdot5y+\left(5y\right)^2\)

\(=9x^2+30xy+25y^2\)

14: \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2\)

\(=4x^2+12xy+9y^2\)

17: \(\left(x^2+9\right)^2=\left(x^2\right)^2+2\cdot x^2\cdot9+9^2=x^4+18x^2+81\)

18: \(\left(2x^2+1\right)^2=\left(2x^2\right)^2+2\cdot2x^2\cdot1+1^2=4x^4+4x^2+1\)

21: \(\left(x+2y^2\right)^2=x^2+2\cdot x\cdot2y^2+\left(2y^2\right)^2=x^2+4xy^2+4y^4\)

22: \(\left(2x+3y^2\right)^2\)

\(=\left(2x\right)^2+2\cdot2x\cdot3y^2+\left(3y^2\right)^2\)

\(=4x^2+12xy^2+9y^4\)

13: \(\left(x-1\right)\left(x+1\right)=x^2+x-x-1=x^2-1\)

14: \(\left(x-5\right)\left(x+5\right)=x^2+5x-5x-25=x^2-25\)

15: \(\left(x-6\right)\left(6+x\right)\)

=(x-6)(x+6)

\(=x^2+6x-6x-36=x^2-36\)

16: \(\left(2x+1\right)\left(2x-1\right)=4x^2-2x+2x-1=4x^2-1\)

17: \(\left(x-2y\right)\left(x+2y\right)=x^2+2xy-2xy-4y^2=x^2-4y^2\)

18: \(\left(5x-3y\right)\cdot\left(3y+5x\right)\)

\(=\left(5x-3y\right)\left(5x+3y\right)\)

\(=25x^2+15xy-15xy-9y^2=25x^2-9y^2\)

19: \(\left(\dfrac{1}{x}-5\right)\left(\dfrac{1}{x}+5\right)=\left(\dfrac{1}{x}\right)^2+\dfrac{5}{x}-\dfrac{5}{x}-25=\dfrac{1}{x^2}-25\)

20: \(\left(x-\dfrac{3}{2}\right)\left(x+\dfrac{3}{2}\right)=x^2+\dfrac{3}{2}x-\dfrac{3}{2}x-\dfrac{9}{4}=x^2-\dfrac{9}{4}\)

21: \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\left(\dfrac{x}{3}\right)^2+\dfrac{xy}{12}-\dfrac{xy}{12}-\left(\dfrac{y}{4}\right)^2\)

\(=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)

22: \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\left(\dfrac{x}{y}\right)^2+\dfrac{2}{3}\cdot\dfrac{x}{y}-\dfrac{2}{3}\cdot\dfrac{x}{y}-\left(\dfrac{2}{3}\right)^2\)

\(=\left(\dfrac{x}{y}\right)^2-\left(\dfrac{2}{3}\right)^2=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)

23: \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)=\left(\dfrac{y}{3}+\dfrac{x}{2}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)\)

\(=\left(\dfrac{y}{3}\right)^2-\dfrac{x}{2}\cdot\dfrac{y}{3}+\dfrac{x}{2}\cdot\dfrac{y}{3}-\left(\dfrac{x}{2}\right)^2\)

\(=\left(\dfrac{y}{3}\right)^2-\left(\dfrac{x}{2}\right)^2=\dfrac{y^2}{9}-\dfrac{x^2}{4}\)

24: \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)\)

\(=4x^2+\dfrac{4}{3}x-\dfrac{4}{3}x-\dfrac{4}{9}=4x^2-\dfrac{4}{9}\)

a, 3.27.9

= 3.3³.3²

= 3¹⁺³⁺²

= 3⁴⁺²

= 3⁶

b; 25.5.125

= 5².5¹.5³

= 5²⁺¹⁺³

= 5³⁺³

= 5⁶

c; 49.7.343

= 7².7¹.7³

= 7²⁺¹⁺³

= 7³⁺³

= 7⁶

d; \(\dfrac{2}{3}\).\(\dfrac{4}{9}\).\(\dfrac{8}{27}\)

=   \(\left(\dfrac{2}{3}\right)\)¹.\(\left(\dfrac{2}{3}\right)\)².\(\left(\dfrac{2}{3}\right)\)³

= \(\left(\dfrac{2}{3}\right)\)¹⁺²⁺³

= \(\left(\dfrac{2}{3}\right)\)³⁺³

= \(\left(\dfrac{2}{3}\right)\)⁶

e; \(\dfrac{3}{4}\).\(\dfrac{9}{16}\).\(\dfrac{27}{64}\)

= \(\left(\dfrac{3}{4}\right)\)¹.\(\left(\dfrac{3}{4}\right)\)².\(\left(\dfrac{3}{4}\right)\)³

=  \(\left(\dfrac{3}{4}\right)\)¹⁺²⁺³

= \(\left(\dfrac{3}{4}\right)\)³⁺³

= \(\left(\dfrac{3}{4}\right)\)⁶

f; \(\dfrac{2}{3}\).\(\dfrac{8}{27}\).\(\dfrac{16}{81}\)

= \(\left(\dfrac{2}{3}\right)\)¹.\(\left(\dfrac{2}{3}\right)\)³.\(\left(\dfrac{2}{3}\right)\)⁴

= \(\left(\dfrac{2}{3}\right)\)¹⁺³⁺⁴

= \(\left(\dfrac{2}{3}\right)\)⁴⁺⁴

= \(\left(\dfrac{2}{3}\right)\)8

                 Giải:

Vì hai bạn cùng thời điểm xuất phát, cùng đến nhà hát vào cùng một lúc nên thời gian đi của hai bạn bằng nhau.

Gọi vận tốc của bạn Lan là \(x\) (km/h); \(x\) > 0

Thời gian bạn Lan  đi đến nhà hát bằng thời gian bạn Điệp đi đến nhà hát và bằng:

                    6 : \(x\) = \(\dfrac{6}{x}\) (giờ)

Vận tốc của bạn Điệp khi đi đến nhà hát là:

                   7 : \(\dfrac{6}{x}\) = \(\dfrac{7}{6}\)\(x\) (km/h)

Theo  bài ra ta có phương trình:

                 \(\dfrac{7}{6}x\) - \(x\) = 2 

              \(x\times\)(\(\dfrac{7}{6}\) - 1) = 2

              \(x\) \(\times\) \(\dfrac{1}{6}\) = 2

              \(x\)     =   2  : \(\dfrac{1}{6}\)

              \(x\)     = 12 

Vậy vận tốc của Lan là 12 km/h

Vận tốc của Điệp là: 12 + 2 = 14 (km/h)

Kết luận: Vận  tốc của Lan 12km/h

              Vận tốc  của Điệp là: 14 km/h

\(\widehat{aOt}=\dfrac{\widehat{aOc}}{2}\)(Ot là phân giác của góc aOc)

\(\widehat{bOz}=\dfrac{\widehat{dOb}}{2}\)(Oz là phân giác của góc dOb)

mà \(\widehat{aOc}=\widehat{bOd}\)(hai góc đối đỉnh)

nên \(\widehat{aOt}=\widehat{bOz}\)

mà \(\widehat{aOt}+\widehat{bOt}=180^0\)(hai góc kề bù)

nên \(\widehat{bOt}+\widehat{bOz}=180^0\)

=>Ot và Oz là hai tia đối nhau

ok bạn nhé mình cảm ơn

bạn cho mik xin hình đc ko?

tui viết ra giấy rồi chụp hình được không?

Sửa đề: \(\left(-1\dfrac{1}{5}\right)\left(-1\dfrac{1}{6}\right)\left(-1\dfrac{1}{7}\right)\left(-1\dfrac{1}{8}\right)\left(-1\dfrac{1}{9}\right)\left(-1\dfrac{1}{10}\right)\)

\(=\dfrac{-6}{5}\cdot\dfrac{-7}{6}\cdot...\cdot\dfrac{-11}{10}\)

\(=\dfrac{6}{5}\cdot\dfrac{7}{6}\cdot...\cdot\dfrac{11}{10}=\dfrac{11}{5}\)

Bài 4:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=k\)

=>\(\left\{{}\begin{matrix}a=bk\\b=ck\\c=ak\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=ak\\b=ak\cdot k=ak^2\\a=ak^2\cdot k=ak^3\end{matrix}\right.\)

=>ak³=a

mà a<>0

nên k³=1

=>k=1

=>\(\left\{{}\begin{matrix}a=b\cdot1=b\\b=c\cdot k=c\\c=a\cdot k=a\cdot1=a\end{matrix}\right.\Leftrightarrow a=b=c\)

\(A=\left(a+2b-3c\right)^{2024}\cdot\left(a+b\right)^{100}\)

\(=\left(a+2a-3a\right)^{2024}\cdot\left(a+a\right)^{100}\)

=0