`@` `\text {Ans}`

`\downarrow`

`32^x \div 16^x = 128`

`=> (2^5)^x \div (2^4)^x = 2^7`

`=> 2^(5x) \div 2^(4x) = 2^7`

`=> 2^(5x - 4x) = 2^7`

`=> 2^x = 2^7`

`=> x = 7`

Vậy, `x = 7.`

x = 7

`32^{x} : 16^{x} = 128`

`=>(2^5)^{x} : (2^{4})^{x} = 2^7`

`=>2^{5x} : 2^{4x} = 2^{7}`

`=>2^{x} = 2^{7}`

`=>x = 7`

Vậy `x=7`

X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128

X x 127/128                                            = 127/128

                                                        X   = 127/128 : 127/128

                                                        X   = 1

Không thích khai triển hằng đẳng thức bậc 5 thì có thể làm thế này, dễ hiểu dễ biến đổi:

\(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-\dfrac{3}{4}sin^22x\)

\(=1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}+\dfrac{3}{8}cos4x\)

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x\)

\(sin^{10}x+cos^{10}x=\left(sin^6x+cos^6x\right)\left(sin^4x+cos^4x\right)-sin^4x.cos^4x\left(sin^2x+cos^2x\right)\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)\left(\dfrac{3}{4}+\dfrac{1}{4}cos4x\right)-\dfrac{1}{16}sin^42x\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}cos^24x-\dfrac{1}{16}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)^2\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)-\dfrac{1}{64}\left(1-2cos4x+cos^24x\right)\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{64}+\dfrac{3}{64}cos8x-\dfrac{1}{64}+\dfrac{1}{32}cos4x-\dfrac{1}{64}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)\)

\(=\dfrac{63}{128}+\dfrac{13}{32}cos4x+\dfrac{5}{128}cos8x\)

=>5<x<=7

=>x\(\in\left\{6;7\right\}\)

Từ đề bài suy ra

$2^5<2^x \leq 2^7$.

$\rightarrow$ $5 < x \leq 7$

$\rightarrow$ $x \in \{6;7\}$

= 0 x (1+2+3+4)

= 0

hok tốt

...........

= 0 nha bạn                                                                                                                                                                                                           học tốt nha

   (128 - 32 x 4) x (1 + 2 + 3 + 4)

= (128 - 128) x 10

=  0 x 10

= 0

32<x<128

=>\(x\in\left\{33;34;35;...;127\right\}\)

\(x=\left\{33;34;35;36;37;38;39;40;...;127;128\right\}\)

128 : 4 + 625 : 125 - 3 x 32

= 32 + 5 - 96

= 37 - 96

= - 59.