`a)`
` x^3 +2x^2 +4x +3 =0`
` => x^3 +x^2 + x^2 +x +3x+3 = 0`
` => x^2(x+1) + x(x+1) +3(x+1) = 0`
` => (x^2 +x+3)(x+1) = 0`
Ta có
` x^2 + x + 3 = (x^2+x+1/4) -1/4 +3 = (x+1/2)^2 +11/4 \ge 11/4 > 0`
nên ` x +1 = 0 => x= -1`
`c)`
` (x-2)(x+2)(x^2-10) = 72`
` => (x^2-4)(x^2 -10) =72`
Đặt ` x^2 =a (a\ge0)`
` => (a-4)(a-10) = 72`
` => a^2 - 14a +40 -72 = 0`
` => a^2 -14a -32=0`
` => a^2 -16a +2a -32 = 0`
` => a(a-16) +2(a-16)= 0`
` => (a+2)(a-16) = 0`
` => a= -2` hoặc ` a= 16`
Mà ` a \ge 0 => a = 16`
` => x^2 = 16 => x = ± 4`
Vậy ` x =±4`
`d)`
` (x+67)/1966 + (x+68)/1965 + (x+69)/(1964) + (x+70)/1963 + (x+2053)/5 = 0`
` => ((x+67)/1966+1) + ((x+68)/1965+1) + ((x+69)/(1964)+1) + ((x+70)/1963+1) +((x+2053)/5 -4) = 0`
` => (x+2033)/1966 + (x+2033)/1965 + (x+2033)/1964 + (x+2033)/1963 + (x+2033)/5 = 0`
` => (x+2033)(1/1966 +1/1965 + 1/1964 + 1/1963 - 1/5) = 0`
Ta thấy ` 1/1966 +1/1965 + 1/1964 + 1/1963 - 1/5 \ne 0`
nên ` x +2033 = 0 => x= -2033`
Vậy ` x= -2033`
