\(C< \frac{2}{3}.\frac{4}{5}......\frac{80}{81}\Rightarrow C.C< \frac{C.2....80}{3.5....81}=\frac{1.2.3....79.80}{2.3.4....81}=\frac{1}{81}=\left(\frac{1}{9}\right)^2mà:C>0\Rightarrow C< \frac{1}{9}\)

Shitbo ơi em có thể giải theo cách cấp 1 được không?

a,10

b,6

làm được hai câu thông cảm

a) \(3^{x-2}=27\cdot9\)

\(3^{x-2}=3^3\cdot3^2=3^5\)

\(\Rightarrow\)\(x-2=5\Rightarrow x=7\)

b) \(2^{x+1}+2^{x+3}=80\)

\(\Rightarrow2^{x+1}\left(1+2^2\right)=80\)

\(\Rightarrow2^{x+1}\cdot5=80\)

\(\Rightarrow2^{x+1}=16=2^4\)

\(\Rightarrow x+1=4\Rightarrow x=3\)

c) \(2^{2x-3}=16\cdot8\)

\(2^{2x-3}=2^4\cdot2^3=2^7\)

\(\Rightarrow2x-3=7\)

\(\Rightarrow2x=4\Rightarrow x=2\)

d) \(2^{x-2}\cdot2^x=64\)

\(\Rightarrow2^{x-2+x}=64=2^6\)

\(\Rightarrow x-2+x=6\)

\(\Rightarrow2x-2=6\)

\(\Rightarrow2x=8\Rightarrow x=4\)

Giải:

a) \(3^{x-2}=27.9\)

\(\Leftrightarrow3^{x-2}=3^3.3^2\)

\(\Leftrightarrow3^{x-2}=3^5\)

Vì \(3=3\)

Nên \(x-2=5\)

\(\Leftrightarrow x=5+2\)

\(\Leftrightarrow x=7\)

Vậy x = 7.

b) \(2^{x+1}+2^{x+3}=80\)

\(\Leftrightarrow2^{x+1}\left(1+2^2\right)=80\)

\(\Leftrightarrow2^{x+1}.5=80\)

\(\Leftrightarrow2^{x+1}=\dfrac{80}{5}=16\)

\(\Leftrightarrow2^{x+1}=2^4\)

Vì \(2=2\)

Nên \(x+1=4\)

\(\Leftrightarrow x=4-1\)

\(\Leftrightarrow x=3\)

Vậy x = 3.

c) \(2^{2x-3}=16.8\)

\(\Leftrightarrow2^{2x-3}=2^4.2^3\)

\(\Leftrightarrow2^{2x-3}=2^7\)

Vì \(2=2\)

Nên \(2x-3=7\)

\(\Leftrightarrow2x=7+3=10\)

\(\Leftrightarrow x=\dfrac{10}{2}=5\)

Vậy x = 5.

d) \(2^{x-2}.2^x=64\)

\(2^{2x-2}=2^6\)

Vì \(2=2\)

Nên \(2x-2=6\)

\(\Leftrightarrow2x=6+2=8\)

\(\Leftrightarrow x=\dfrac{8}{2}=4\)

Vậy x = 4.

Chúc bạn học tốt!

Thiếu đề bạn ơi có mỗi chữ tính thì làm dc j

Đề???????????

mình cần gấp giúp mình với

b) 52-\(|\)x\(|\)=-80

         \(|\)x\(|\)=52-(-80)

         \(|\)x\(|\)=52+80

         \(|\)x\(|\)=132

    Vậy x=-132

Ta có: \(\frac{2x-9}{240}=\frac{39}{80}=\frac{117}{240}\)

=>\(2x-9=117\)

\(2x=117+9=126\)

\(x=126:2=63\)

Ta có :\(\frac{39}{80}=\frac{117}{240}\)

nên \(\frac{2x-9}{240}=\frac{117}{240}\)

\(\Rightarrow2x-9=117\)

\(2x=117+9\)

\(2x=126\)

\(x=126:2\)

\(x=63\)

a) \(15+2\left|x\right|=-3\\ \\ < =>2\left|x\right|=15-\left(-3\right)\\ < =>2\left|x\right|=18\\ =>\left|x\right|=\frac{18}{2}=9\\ =>x=9hoặcx=-9\)

b) \(\left|x-2\right|=7\\ < =>x-2=7hoặcx-2=-7\\ =>x=9hoặcx=-5\)

c) \(100-4.x^2=224\\ < =>4.x^2=100-224=-124\\ < =>x^2=-\frac{124}{4}=-31\\ Mà:x^2\ge0\\ =>xkhôngcógiátrịnàothoảmãn\)

d)\(2x-\frac{9}{240}=\frac{39}{80}\\ < =>2x-\frac{3}{80}=\frac{39}{80}\\ =>2x=\frac{39}{80}+\frac{3}{80}=\frac{21}{40}\\ =>x=\frac{\frac{21}{40}}{2}=\frac{21}{80}\)

\(15+2\left|x\right|=-3\)

\(\Rightarrow2\left|x\right|=15+3\)

\(\Rightarrow2\left|x\right|=18\)

\(\Rightarrow\left|x\right|=\frac{18}{2}\)

\(\Rightarrow\left|x\right|=9\)

\(\Rightarrow\left[\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

Vậy, x = 9 hoặc x = -9.

\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\)\(\Rightarrow\)\(2x-9\)=\(\dfrac{240.39}{80}\)=\(117\)

\(2x-9=117\)\(\Rightarrow\)\(2x=117+9=126\)\(\Rightarrow\)\(x=126:2=63\)

\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\) \(\Rightarrow\) \(2x-9=\)\(\dfrac{240.39}{80}\)=117

\(2x-9=117\)\(\Rightarrow\)\(2x=117-9=108\)\(\Rightarrow\)\(x=108:2=54\)