Question

Tìm x bt :

a) x + x - 1 + x - 2 +...+z - 50 = 255

b) \(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)

c) -2^x + 624 = 5^y (x,y \(\in\) N )

d) \(\dfrac{x+5}{2}=\dfrac{8}{x+5}\)

e) x + ( x + 1 ) + ( x + 2 ) +...+ ( x + 30 ) =1240

f) ( x - 1 ) + ( x - 2 ) + ... + (x - 19 ) + ( x - 20 ) = -610

g) ( 7x - 11 )³ = ( -3 )² . 15 + 208

Answer 1

a)

\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)

e)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)

f)

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)