\(\left(2-3x\right)\left(7+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2-3x=0\\7+2x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=2\\2x=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{7}{2}\end{cases}}\)

Bài làm :

\(\left(2-3x\right)\left(7+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2-3x=0\\7+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-7}{2}\end{cases}}\)

Vậy \(x=\frac{2}{3}\) hoặc \(x=\frac{-7}{2}\) .

Học tốt nhé

\(2^{x+1}.3^y=12^x\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\Leftrightarrow\frac{2^{2x}.3^x}{2^{x+1}.3^y}=1\)

\(\Leftrightarrow2^{x-1}.3^{x-y}=1\)

Vì \(2^{x-1}\ge1,3^{x-y}\ge1\)mà đề yêu cầu giải dấu "=" xảy ra, khi đó:

\(\hept{\begin{cases}2^{x-1}=1\\3^{x-y}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\end{cases}}\)

Đặt x = y + k (vì x - y > 0 ; k > 0)

Ta có 2^x - 2^y = 256

=> 2^{y + k} - 2^y = 256

= 2^y(2^k - 1) = 256

Vì y > 0

=> 2^y là số chẵn

Lại có k > 0

=> 2^k  chẵn 

=> 2^k - 1 lẻ 

Nếu 2^k - 1 = 1

=> 2^k = 2

=> k = 1(tm)

=> y = 9 => x = 10

Do 2^k - 1 lẻ mà 1 ước lẻ duy nhất của 256 

=> Không tồn tại số 2^k - 1 > 1 là ước của 256

Vậy y = 9 ; x  = 10

Bài làm :

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x.\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Học tốt

\(3.3^{x-2}+5.3^{x-1}=162\)

\(\Rightarrow3^{x-1}+5.3^{x-1}=162\)

\(\Rightarrow3^{x-1}.6=162\)

\(\Rightarrow3^{x-1}=27=3^3\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x=4\)

\(3\cdot3^{x-2}+5\cdot3^{x-1}=162\) 

\(3^{x-1}+5\cdot3^{x-1}=162\) 

\(3^{x-1}\left(5+1\right)=162\) 

\(3^{x-1}\cdot6=162\) 

\(3^{x-1}=27\)  

\(3^{x-1}=3^3\) 

\(x-1=3\) 

\(x=4\) 

\(5S=5+1+\frac{1}{5^2}+...+\)\(\frac{1}{5^{2019}}\)

\(5S-S=4S=5-\frac{1}{5^{2019}}\)

\(\Rightarrow S=\frac{\frac{5^{2020}}{5^{2019}}}{4}\)

\(S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\)

\(5S=5\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)

\(5S=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)

\(5S-S=4S\)

\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)

\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-1-\frac{1}{5}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2020}}\)

\(=5-\frac{1}{5^{2020}}\)

\(4S=5-\frac{1}{5^{2020}}\Rightarrow S=\frac{5-\frac{1}{5^{2020}}}{4}\)

4⁵ . 9⁴ - 2 . 6⁹/2¹⁰ . 3⁸ + 6⁸ . 20\(=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)

Có thể

KHÔNG THỂ

TH1:\(x\ge-2\Rightarrow\left|x+2\right|=x+2;\left|x+5\right|=x+5\)

\(\Rightarrow x+2+x+5=x\)

\(\Rightarrow x=-7\text{(loại vì ko thỏa mãn đk x\ge2 đang xét)}\)

TH2:\(-5\le x< -2\Rightarrow\hept{\begin{cases}\left|x+2\right|=-\left(x+2\right)=-x-2\\\left|x+5\right|=x+5\end{cases}}\)

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

\(\Rightarrow3=x\text{ hay }x=3\left(\text{loại}\right)\)

TH3:\(x< -5\Rightarrow\hept{\begin{cases}\left|x+2\right|=-x-2\\\left|x+5\right|=-x-5\end{cases}}\)

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

\(\Rightarrow-3x=7\)

\(\Rightarrow x=\frac{-7}{3}\left(\text{loại}\right)\)

\(\text{Vậy phương trình vô nghiệm}.\)