Bài 2:     

a) 2(5x -8) –( x² +10x) = -17

=>  10x - 16 – x² - 10x = -17

=>  - 16 – x²  = -17

=>    x²  =  - 16 +17

=>  x²   = 1 

=>     \(\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

 b)  x²  -3x - 4 = 0

=>    x²  - 4x + x - 4 = 0

=>   ( x²  - 4x ) + ( x - 4 ) = 0

=>   x ( x  - 4 ) + ( x - 4 ) = 0

=>( x  - 4 )( x + 1 ) = 0

=>  \(\orbr{\begin{cases}x-4=0\\x+1=0\end{cases}}\)        

=>  \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)        

c)     x(2x -1) + 2 x² = 3

=>     2x²  - x  + 2 x² = 3

=>     4x²  - x  - 3 = 0

=>  4x²  - 4x  +3x  - 3 = 0

=> ( 4x²  - 4x ) + ( 3x  - 3 ) = 0

=>   4x( x - 1 ) + 3( x  - 1 ) = 0

=> ( x - 1 )( 4x + 3) = 0

=>  \(\orbr{\begin{cases}x-1=0\\4x+3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{4}\end{cases}}\)         

A= 3x³ - (3x -2)x²  - 2x(x+1)

A= 3x³ - 3x³ + 2x² - 2x² -2x

A= -2x

Thay x =-20 vào A ta được:

A = -2.(-20) = 40

Vậy A= 40 khi x = -20 

b) C= x(2x+1) - x²(x+2) + x³ -x + 3

C= 2x² + x - x³ - 2x² + x³ -x +3

C= (2x² - 2x²) + (x-x) - (x³ -x³) +3 

C = 3

Vậy C= 3

đây nè bạn

d) \(\frac{2^{2x-3}}{4^{10}}\)\(=8^3.16^5\)

\(\frac{2^{2x-3}}{\left(2^2\right)^{10}}\)\(=\left(2^3\right)^3.\left(2^4\right)^5\)

\(\frac{2^{2x-3}}{2^{20}}=2^9.2^{20}\)

\(2^{2x-3}=2^{29}.2^{20}\)

\(2^{2x-3}=2^{49}\)

=> 2x-3=49

=> 2x= 52

=> x=26

Vậy x= 26

\(\frac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)   

\(\frac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)   

\(\frac{2^{2x-5}}{2^{20}}=2^9\cdot2^{20}\)   

\(2^{2x-5}=2^{49}\)   

2x - 5 = 49 

2x = 54 

x = 27 

\(\left(-2\right)^x=4^5\cdot16^2\)   

\(\left(-2\right)^x=\left(2^2\right)^5\cdot\left(2^4\right)^2\)   

\(\left(-2\right)^x=2^{10}\cdot2^8\)   

\(\left(-2\right)^x=2^{18}\)   

\(\Rightarrow x=18\)

\(\frac{2^{4-x}}{16^5}=32^6\)   

\(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)   

\(\frac{2^{4-x}}{2^{20}}=2^{30}\)   

\(2^{4-x}=2^{30}\cdot2^{20}\)   

\(2^{4-x}=2^{50}\)   

\(4-x=50\)   

\(x=4-50\)   

\(x=-46\)   

\(3^x=\frac{9^4}{27^3}\)   

\(3^x=\frac{\left(3^2\right)^4}{\left(3^3\right)^3}\)   

\(3^x=\frac{3^8}{3^9}\)   

\(3^x=3^{-1}\)   

x = -1