\(\frac{45^{10}.5^{20}}{75^{15}}\)

\(=\frac{\left(15.3\right)^{10}.5^{20}}{\left(15.5\right)^{15}}\)

\(=\frac{15^{10}.3^{10}.5^{20}}{15^{15}.5^{15}}\)

\(=\frac{3^{10}.5^5}{15^5}=\frac{3^{10}.5^5}{3^5.5^5}=3^5=243\)

\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(9.5\right)^{10}.5^{20}}{\left(3.5.5\right)^{15}}=\frac{9^{10}.5^{10}.5^{20}}{3^{15}.5^{15}.5^{15}}=\frac{9^{10}.5^{30}}{3^{15}.5^{30}}=\frac{9^{10}}{3^{15}}=243\)

\(4\left(2x-5\right)^{100}=9\left(5-2x\right)^{98}\)

\(\Leftrightarrow4\left(5-2x\right)^{100}=9\left(5-2x\right)^{98}\)

\(\Leftrightarrow4\left(5-2x\right)^2=9\)

\(\Leftrightarrow4\left(5-2x\right)^2-9=0\)

\(\Leftrightarrow[2\left(5-2x\right)-3][2\left(5-2x\right)+3]=0\)

\(\Leftrightarrow\left(10-4x-3\right)\left(10-4x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7-4x=0\\13-4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{13}{4}\end{cases}}}\)

Ta có: \(xy+x-y=2\)

\(\Leftrightarrow\left(xy+x\right)-\left(y+1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=1=1.1=\left(-1\right).\left(-1\right)\)

Nếu x,y nguyên thì ta mới tìm được nhé

+ Nếu: \(\hept{\begin{cases}x-1=1\\y+1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x-1=-1\\y+1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-2\end{cases}}\)

\(\frac{15^{10}.5^{10}}{75^{10}}\)

\(=\frac{15^{10}.5^{10}}{\left(15.5\right)^{10}}\)

\(=\frac{15^{10}.5^{10}}{15^{10}.5^{10}}\)

\(=1\)

          Bài làm :

Ta có :

\(\frac{15^{10}\times5^{10}}{75^{10}}=\frac{\left(15\times5\right)^{10}}{75^{10}}=\frac{75^{10}}{75^{10}}=1\)

nhanh giúp em với ạ

\(\frac{x-2}{3}=\frac{4}{5}\)

\(\Leftrightarrow x-2=\frac{12}{5}\)

\(\Leftrightarrow x=\frac{22}{5}\)

ĐKXĐ : \(x\ne4\)

\(\frac{2x-3}{4-x}>0\)

+) TH1 : \(\hept{\begin{cases}2x-3>0\\4-x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x< 4\end{cases}\Leftrightarrow}\frac{3}{2}< x< 4}\)

+) TH2 : \(\hept{\begin{cases}2x-3< 0\\4-x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< \frac{3}{2}\\x>4\end{cases}\left(ktm\right)}}\)

Vậy với \(\frac{3}{2}< x< 4\)thì \(\frac{2x-3}{4-x}>0\)

\(\frac{2x-3}{4-x}>0\)

ĐK : x khác 4

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-3>0\\4-x>0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x>3\\-x>-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x< 4\end{cases}}\Leftrightarrow\frac{3}{2}< x< 4\)

2. \(\hept{\begin{cases}2x-3< 0\\4-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x< 3\\-x< -4\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{3}{2}\\x>4\end{cases}}\)( loại )

=> Với \(\frac{3}{2}< x< 4\)thì thỏa mãn đề bài

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)

\(2A=3A-A\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)

\(=1-\frac{1}{3^{2008}}\)

\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)