Giúp mik giải các pt sau với a, 2(x-3)-1/2(x+4)=5x-3 b, 3 – 5/x-1 + 4/3x^2 – 6x + 3 = 0 c, 3x^2 – 8x + 4=0 d, 1+8x/4+8x – 4x/12x-6 + 32x^2/ 3(4-16x^2)

Đáp án: a.$x=-\dfrac{10}{7}$

                b.$ x\in\{\dfrac73,\dfrac43\}$ 

                c.$x\in\{2,\dfrac23\}$

                d.$x\approx \:0.06262\dots $

Giải thích các bước giải:

a.ĐKXĐ : $x\ne -4$

Ta có :
$2(x-3)-\dfrac12(x+4)=5x-3$

$\to \dfrac{3}{2}x-8=5x-3$

$\to -\dfrac{7}{2}x=5$

$\to x=-\dfrac{10}{7}$

b.ĐKDXĐ $x\ne 1$

Ta có :
$3-\dfrac{5}{x-1}+\dfrac{4}{3x^2-6x+3}=0$

$\to 3-\dfrac{5}{x-1}+\dfrac{4}{3(x^2-2x+1)}=0$

$\to 3-\dfrac{5}{x-1}+\dfrac{4}{3(x-1)^2}=0$

$\to 9\left(x-1\right)^2-15\left(x-1\right)+4=0$

$\to 9x^2-33x+28=0$

$\to (3x-7)(3x-4)=0$

$\to x\in\{\dfrac73,\dfrac43\}$

c.$3x^2-8x+4=0$

$\to 3x^2-6x-2x+4=0$

$\to 3x(x-2)-2(x-2)=0$

$\to (3x-2)(x-2)=0$

$\to x\in\{2,\dfrac23\}$

d.ĐKXĐ: $x\ne 2,\pm\dfrac12$

Ta có:
$\dfrac{1+8x}{4+8x}-\dfrac{4}{12-6x}+\dfrac{32x^2}{3(4-16x^2)}=0$

$\to \dfrac{1+8x}{4(2x+1)}+\dfrac{2}{3(x-2)}-\dfrac{8x^2}{3(4x^2-1)}=0$

$\to \dfrac{1+8x}{4(2x+1)}+\dfrac{2}{3(x-2)}-\dfrac{8x^2}{3(2x-1)(2x+1)}=0$

$\to 3\left(8x+1\right)\left(x-2\right)\left(2x-1\right)+8\left(2x+1\right)\left(2x-1\right)-32x^2\left(x-2\right)=0$

$\to 16x^3-18x^2+33x-2=0$

$\to x\approx \:0.06262\dots $