Đáp án:
Bài 1:
a.$ x\in\{2, -5\}$
b.$x=-\dfrac1{13}$
c.Vô nghiệm
Bài 2:
a.$x\le 0$
b.$-\dfrac{15}6<x\le 1$
Giải thích các bước giải:
Bài 1:
a.Ta có:
$x(x+1)(x+2)(x+3)=120$
$\to x(x+3)\cdot (x+1)(x+2)=120$
$\to (x^2+3x)\cdot (x^2+3x+2)=120$
$\to (x^2+3x)^2+2(x^2+3x)=120$
$\to (x^2+3x)^2+2(x^2+3x)+1=121$
$\to (x^2+3x+1)^2=121$
$\to x^2+3x+1=11$ hoặc $x^2+3x+1=-11$
$\to x\in\{2, -5\}$
b.ĐKXĐ: $x\ne -\dfrac13, \dfrac13$
Ta có:
$\dfrac{4x-1}{6x-2}-\dfrac{2x+3}{3x+1}=\dfrac{1-26x}{9x^2-1}$
$\to\dfrac{4x-1}{2(3x-1)}-\dfrac{2x+3}{3x+1}=\dfrac{1-26x}{(3x-1)(3x+1)}$
$\to \dfrac{4x-1}{2\left(3x-1\right)}\cdot \:2\left(3x-1\right)\left(3x+1\right)-\dfrac{2x+3}{3x+1}\cdot \:2\left(3x-1\right)\left(3x+1\right)=\dfrac{1-26x}{\left(3x-1\right)\left(3x+1\right)}\cdot \:2\left(3x-1\right)(3x+1)$
$\to \left(4x-1\right)\left(3x+1\right)-2\left(2x+3\right)\left(3x-1\right)=2\left(1-26x\right)$
$\to -13x+5=2-52x$
$\to 39x=-3$
$\to x=-\dfrac1{13}$
c.ĐKXĐ: $x\ne 1$
Ta có:
$\dfrac1{x-1}+\dfrac{2x^2+5}{x^3-1}=\dfrac{4}{x^2+x+1}$
$\to \dfrac{x^2+x+1}{(x-1)(x^2+x+1)}+\dfrac{2x^2+5}{(x-1)(x^2+x+1)}=\dfrac{4(x-1)}{(x-1)(x^2+x+1)}$
$\to x^2+x+1+2x^2+5=4(x-1)$
$\to 3x^2+x+6=4\left(x-1\right)$
$\to 3x^2-3x+10=0$
Mà $3x^2-3x+10=3(x-\dfrac12)^2+\dfrac{27}4>0$
$\to$Phương trình vô nghiệm
Bài 2:
a.Ta có:
$\dfrac{2x}{5}+\dfrac{3-2x}{3}\ge \dfrac{3x+2}{2}$
$\to \dfrac{2x}{5}\cdot \:30+\dfrac{3-2x}{3}\cdot \:30\ge \dfrac{3x+2}{2}\cdot \:30$
$\to 12x+10\left(-2x+3\right)\ge \:15\left(3x+2\right)$
$\to -8x+30\ge \:45x+30$
$\to -53x\ge \:0$
$\to x\le 0(1)$
Lại có:
$\dfrac{x}2+\dfrac{3-2x}{5}\ge\dfrac{3x-5}{6}$
$\to \dfrac{x}{2}\cdot \:30+\dfrac{3-2x}{5}\cdot \:30\ge \dfrac{3x-5}{6}\cdot \:30$
$\to 15x+6\left(-2x+3\right)\ge \:5\left(3x-5\right)$
$\to 3x+18\ge \:15x-25$
$\to 12x\le 43$
$\to x\le\dfrac{43}{12}(2)$
Từ $(1), (2)\to x\le 0$
b.Ta có:
$2(3x-4)<3(4x-3)+16$
$\to 6x-8<12x+7$
$\to -6x<15$
$\to x>-\dfrac{15}6(1)$
Lại có:
$4(1+x)\le 3x+5$
$\to 4+4x\le \:3x+5$
$\to x\le 1(2)$
Từ $(1), (2)\to -\dfrac{15}6<x\le 1$
