Lời giải:

$3^{2x+2}=9^{x+3}$

$3^{2x+2}=(3^2)^{x+3}=3^{2(x+3)}=3^{2x+6}$

$\Rightarrow 2x+2=2x+6$

$\Rightarrow 2=6$ (vô lý)

Vậy không có x thỏa mãn đề.

gíup t vs ạ^^

b, \(\dfrac{3}{1\times2}\) + \(\dfrac{3}{2\times3}\)+ \(\dfrac{3}{3\times4}\)+ ...+ \(\dfrac{3}{99\times100}\)

= \(3\) \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{99\times100}\)

= 3 \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))

= 3 \(\times\) ( 1 - \(\dfrac{1}{100}\))

= 3 \(\times\) \(\dfrac{99}{100}\)

= \(\dfrac{297}{100}\)

c, \(\dfrac{3}{1\times2\times2}\) + \(\dfrac{3}{2\times3\times2}\)+ \(\dfrac{3}{3\times4\times2}\)+...+ \(\dfrac{3}{99\times100\times2}\)

= \(\dfrac{3}{2}\) \(\times\)( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\) +...+ \(\dfrac{1}{99\times100}\))

= \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{99}\)- \(\dfrac{1}{100}\))

= \(\dfrac{3}{2}\) \(\times\) (\(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))

= \(\dfrac{3}{2}\) \(\times\) \(\dfrac{99}{100}\)

= \(\dfrac{297}{200}\)

a, 2¹.5².17 = 2.25.17 = 50.17 = 850 

b, 2² + 2³ + 2⁴ = 4 + 8 + 16 = 28

c, 2⁵.3 + 2⁴:8 + 50: 5²

= 32.3 + 16:8 + 50:25

=96 + 2 + 2

= 100

d, 11² - 10² - 3²

= 121 - 100 - 9

= 21 - 9

= 12

e, 1³ + 2³ + 3³ + 4³ + 5³

= ( 1+ 2+3+4+5)²

= 15²

= 225

\(x\) +  \(\dfrac{1}{4}\) + \(x\) + \(\dfrac{3}{2}\) = \(x\) + \(\dfrac{2}{3}\) + \(x\) + 4

\(x\) + \(x\) - \(x\) - \(x\)      = \(\dfrac{2}{3}\) + 4 - \(\dfrac{3}{2}\) - \(\dfrac{1}{4}\)

                      0 = \(\dfrac{35}{12}\) (vô lý)

vậy \(x\) \(\in\) \(\varnothing\)

Dạ cảm ơn cô

a) \(B\left(16\right)=\left\{0;16;32;48;64;80;96\right\}\)

b) \(U\left(135\right)=\left\{1;3;5;9;15;27;45\right\}\)

c) \(B\left(17\right)=\left\{0;17;34;51;68;85\right\}\)

d) \(U\left(75\right)=\left\{1;3;5;15;25;75\right\}\)

e) \(B\left(33\right)=\left\{0;33;66\right\}\)

f) \(U\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

Đề bài là gì vậy bạn>

\(a\times100\) + 50 + \(c\) = \(\overline{a5c}\)

\(a\times100\) +  \(b\times10\) + 6 = \(\overline{ab6}\)

\(745-x=304\)

          \(x=745-304\)

          \(x=441\)

\(...\Rightarrow x+x+\dfrac{x}{43}+\dfrac{x}{8}=14+148+\dfrac{10}{30}+\dfrac{5}{95}\)

\(\Rightarrow\left(1+1+\dfrac{1}{43}+\dfrac{1}{8}\right)x=162+\dfrac{1}{3}+\dfrac{1}{19}\)

\(\Rightarrow\left(\dfrac{2.43.8}{43.8}+\dfrac{1.8}{43.8}+\dfrac{1.43}{43.8}\right)x=\dfrac{162.3.19}{3.19}+\dfrac{1.19}{3.19}+\dfrac{1.3}{19.3}\)

\(\Rightarrow\left(\dfrac{688}{344}+\dfrac{8}{344}+\dfrac{43}{344}\right)x=\dfrac{9234}{57}+\dfrac{19}{57}+\dfrac{3}{57}\)

\(\Rightarrow\dfrac{739}{344}x=\dfrac{9256}{57}\)

\(\Rightarrow x=\dfrac{9256}{57}:\dfrac{739}{344}=\dfrac{9256}{57}.\dfrac{344}{739}=\dfrac{\text{3184064}}{\text{42123}}\)

\(x\) + \(\dfrac{1}{10}\) + \(x\) + \(\dfrac{1}{11}\) = \(x\) + \(\dfrac{1}{21}\)

\(x+x-x\) = \(\dfrac{1}{21}\) - \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)

\(x\)             = -  \(\dfrac{331}{2310}\)

\(x+\dfrac{1}{10}+x+\dfrac{1}{11}=x+\dfrac{1}{21}\)

\(x+x-x=\dfrac{1}{21}-\dfrac{1}{10}-\dfrac{1}{11}\)

\(x=\dfrac{-11}{210}-\dfrac{1}{11}\)

\(x=\dfrac{-331}{2310}\)

f)

`(2x+1)^3=343`

`(2x+1)^3=7^3`

`=>2x+1=7`

`2x=7-1`

`2x=6`

`x=6:2`

`x=3`

g)

`(x-1)^3 =125`

`(x-1)^3 =5^3`

`=>x-1=5`

`x=6`

h)

`2^(x+2)-2^x=96`

`2^x *2^2 -2^x =96`

`2^x (2^2 -1)=96`

`2^x *3=96`

`2^x =32`

`2^x =2^5`

`=>x=5`

i)

`(x-5)^4 =(x-5)^6` (`x>=5`)

`(x-5)^6 -(x-5)^4 =0`

`(x-5)^4 [(x-5)^2 -1]=0`

`=>x-5=0` hoặc `(x-5)^2 -1=0`

`<=>x=5` hoặc `(x-5)^2 =1`

`<=>x=5` hoặc `x-5=1` hoặc `x-5=-1`

`<=>x=5` hoặc `x=6` hoặc `x=4`

j)

`720:[41-(2x-5)]=2^3 *5`

`720:[41-(2x-5)]=8*5`

`720:[41-(2x-5)]=40`

`41-(2x-5)=720:40`

`41-(2x-5)=18`

`2x-5=41-18`

`2x-5=23`

`2x=28`

`x=14`