a) 10² và 90^ 10

Ta có : 90¹⁰ = (90⁵)²

Vì 90⁵ > 10 => 90^10 > 10^2

b) (-5)^30 và (-3)^50

Ta có : (-5)^30= (-5^3)^10= -125^10

            (-3)^50= (-3^5)^ 10= -243^10

Vì -125>-243 => (-3)^50 < (-5)^30

c) (-1)^10/16 và 1^50/2

Ta có: (-1)^10/16 = 1/16 = 1/2^4= 2^(-4)

          1^50/2 = 1/2= 2^(-1)

Vì 2^(-1) < 2^(-4) => 1^50/2 < (-1)^10/16

a) \(\left|x-\frac{3}{5}\right|=2x-\frac{2}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{5}=2x-\frac{2}{5}\\x-\frac{3}{5}=\frac{2}{5}-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\3x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{5};\frac{1}{3}\right\}\)

b) \(\left|x+0,37\right|=\left|2x-0,63\right|\)

\(\Leftrightarrow\orbr{\begin{cases}x+0,37=2x-0,63\\x+0,37=0,63-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\3x=0,26\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{13}{150}\end{cases}}\)

Vậy \(x\in\left\{\frac{13}{150};1\right\}\)

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

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

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

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

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

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

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

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

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

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

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

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

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