Đáp án+Giải thích các bước giải:
`b) (2x-3)^2 = 4x+6`
`<=> 4x^2 - 12x + 9 - 4x - 6=0`
`<=> 4x^2 - 16x +3=0`
`<=> (2x)^2 - 2.2x . 4 + 16 -13=0`
`<=> (2x-4)^2 = 13`
`<=> (2x-4)^2 = (±\sqrt{13})^2`
`<=> 2x - 4 =±\sqrt{13}`
`<=> 2x = ±\sqrt{13}+4`
`<=> x = ±(\sqrt{13})/2 + 2`
`d) 2x(x-25)+(x+25).3=0`
`<=> 2x^2 - 50x +3x + 75 =0`
`<=> 2x^2 - 47x + 75 =0`
`<=> 2(x^2 - (47)/2x +(75)/2)=0`
`<=> x^2 - (47)/2x +(75)/2 =0`
`<=> x^2 -2.x (47)/4 +(2209)/(16) - (1609)/(16)=0`
`<=> (x-(47)/4)^2 = (1609)/(16)`
`<=> (x-(47)/4)^2 = (±(\sqrt{1609})/4)^2`
`<=> x-(47)/4 = ±(\sqrt{1609})/4`
`<=> x = ±(\sqrt{1609})/4 + (47)/4`
`<=> x = (±\sqrt{1609} +47)/4`
Vậy `x in {{±\sqrt{1609} +47)/4}`
`f) (x^2+3)(x^2-9)=0`
`<=> x^2 - 9 = 0(` vì `x^2 + 3 ≥ 3 > 0 ∀ x in RR)`
`<=> (x+3)(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x+3=0\\x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-3\\x=3\end{array} \right.\)
Vậy `x in {±3}`
`h) (x+1)^2 + (y-2)^2=0`
`<=>`\(\left[ \begin{array}{l}x+1=0\\y-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\y=2\end{array} \right.\)
Vậy `x = -1; y=2`
