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 $
