\(3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(\Rightarrow x+1=3\)

\(x=3-1\)

\(x=2\)

Vậy x = 2.

\(#Paciupibijd\)

\(3^{x+1}=27\)

\(\Rightarrow3^{x+1}=3^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=3-1\)

\(\Rightarrow x=2\)

TH1: `2<=x<=3` 

\(\left(2x-4\right)+\left(3-x\right)=2x\\ =>2x-4+3-x=2x\\ =>x-1=2x\\ =>2x-x=-1\\ =>x=-1\left(ktm\right)\)

TH2: `x>3` 

\(\left(2x-4\right)-\left(3-x\right)=2x\\ =>2x-4-3+x=2x\\ =>3x-7=2x\\ =>3x-2x=7\\ =>x=7\left(tm\right)\)

TH3: `x<2` 

\(-\left(2x-4\right)+\left(3-x\right)=2x\\ =>-2x+4+3-x=2x\\ =>-3x+7=2x\\ =>2x+3x=7\\ =>5x=7\\ =>x=\dfrac{7}{5}\left(tm\right)\)

Vậy: ...

\(\left|2x-4\right|+\left|3-x\right|=2x\)

Ta có : \(\left|2x-4\right|+\left|3-x\right|\ge\left|2x-4+3-x\right|=\left|x-1\right|\)

\(\Rightarrow\left|x-1\right|=2x\)

\(\)\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x-1=2x\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}2x\ge0\\x-1=-2x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x=-1\end{matrix}\right.\) (loại) hay \(\left\{{}\begin{matrix}x\ge0\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{3}\)

Chứng minh cái gì? chắc là so sánh

\(7^{714}< 8^{714}\)

\(2^{1999}>2^{1998}=\left(2^3\right)^{666}=8^{666}\)

\(\Rightarrow2^{1999}>8^{666}>8^{714}>7^{714}\)

\(A=2+2^2+2^3+...+2^{12}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{10}+2^{11}+2^{12}\right)\\ =\left(2+2^2+2^3\right)+2^3\cdot\left(2+2^2+2^3\right)+...+2^9\cdot\left(2+2^2+2^3\right)\\ =\left(2+4+8\right)+2^3\cdot\left(2+4+8\right)+...+2^9\cdot\left(2+4+8\right)\\ =14+2^3\cdot14+...+2^9\cdot14\\ =14\cdot\left(1+2^3+...+2^9\right)\) 

=> A chia hết cho 14

\(\dfrac{72^3.54^3:8^3}{108^5:4^5}\)

= \(\dfrac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^3:\left(2^3\right)^3}{\left(2^2.3^3\right)^5:\left(2^2\right)^5}\)

= \(\dfrac{2^9.3^6.2^3.3^9:2^9}{2^{10}.3^{15}:2^{10}}\)

= \(\dfrac{3^{15}.2^3}{3^{15}}\)

= `2^3`

Em cảm ơn ạ

a)
 \(x+\dfrac{5}{6}=7,5\\ \Rightarrow x+\dfrac{5}{6}=\dfrac{15}{2}\\ \Rightarrow x=\dfrac{15}{2}-\dfrac{5}{6}\\ \Rightarrow x=\dfrac{20}{3}\)
Vậy...

b) 
\(x.\dfrac{3}{5}-1\dfrac{1}{2}=\dfrac{5}{6}\\ \Rightarrow x.\dfrac{3}{5}-\dfrac{3}{2}=\dfrac{5}{6}\\ \Rightarrow x.\dfrac{3}{5}=\dfrac{5}{6}+\dfrac{3}{2}\\ \Rightarrow x.\dfrac{3}{5}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}:\dfrac{3}{5}\\ \Rightarrow x=\dfrac{35}{9}\)
Vậy...

c) Đề bài thiếu (\(x:\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{3}=?\))

d)
 \(12,5-x=6,75\\ \Rightarrow\dfrac{25}{2}-x=\dfrac{27}{4}\\ \Rightarrow x=\dfrac{25}{2}-\dfrac{27}{4}\\ \Rightarrow x=\dfrac{23}{4}\)
Vậy...

`D = 500 - {5 . [409 - (2^3 - 21)^2] - 1724}`

`D = 500 - {5 . [409 - (8 - 21)^2] - 1724}`

`D = 500 - {5 . [409 - (-13)^2] - 1724}`

`D = 500 - {5 . [409 - 169] - 1724}`

`D = 500 - {5 . 240 - 1724}`

`D = 500 - {1200 - 1724}`

`D = 500 - (-524)`

`D = 500 + 524`

`D = 1024`

`A = 99-97 + 95 - 94 + ... + 3 - 1`

Biểu thức A có số số hạng là: 

`(99 - 1) : 2 + 1 = 50` (số hạng)

`A= (99-97) + (95 - 94) + ... + (3 - 1) `

Biểu thức A có số số nhóm tạo được là: 

`50 : 2 = 25` (nhóm)   

`A = 2 + 2 + ... + 2`

`A = 50 . 2`

`A = 100`

`2520cm^2 = 25,20dm^2 = 0,25m^2`

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-> Các đơn vị đo diện tích cao hơn gấp `100` lần đơn vị đo diện tích bé hơn liền sau