1,5(x+3)-4x(3+x)=0 2,4x(x-2004)-x+2004=0 3,x²+8x+16=0 4,4x²-12x=-9

Đáp án:

 

Giải thích các bước giải:

`1) 5.(x + 3) – 4x.(3 + x) = 0.`

`⇒ 5.(x + 3) – 4x.(x + 3) = 0.`

`⇒ (5 – 4x).(x + 3) = 0.`

`⇒` \(\left[ \begin{array}{l}5-4x=0.\\x+3=0.\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}4x=5.\\x=-3.\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x=\dfrac{5}{4}.\\x=-3.\end{array} \right.\)

Vậy `x=5/4` hoặc `x=-3.`

`2) 4x.(x – 2004) – x + 2004 = 0.`

`⇒ 4x.(x – 2004) – (x – 2004) = 0.`

`⇒ (4x – 1).(x – 2004) = 0.`

`⇒` \(\left[ \begin{array}{l}4x-1=0.\\x-2004=0.\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}4x=1.\\x=2004.\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{4}.\\x=2004.\end{array} \right.\)

Vậy `x = 1/4` hoặc `x = 2004.`

`3) x^2 + 8x + 16 = 0.`

`⇒ x^2 + 2.x.4 + 4^2 = 0.`

`⇒ (x + 4)^2 = 0.`

`⇒ x + 4 = 0.`

`⇒ x = -4.`

Vậy `x = -4.`

`4) 4x^2 – 12x = -9.`

`⇒ 4x^2 – 12x + 9 = 0.`

`⇒ (2x)^2 – 2.2x.3 + 3^2 = 0.`

`⇒ (2x – 3)^2 = 0.`

`⇒ 2x – 3 = 0.`

`⇒ 2x = 3.`

`⇒ x = 3/2.`

$\text{Vậy x = $\dfrac{3}{2}$.}$

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