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3x.|x+1|−2x|x+2|=12
Với x < -2 ta có: 3x.(-x-1)-2x(-x-2)-12=0
<=> -3x2 - 3x + 2x2 + 4x -12 =0
<=> -x2 - x - 12=0
$\Leftrightarrow $ -(x2 +x+12)=0 ( vô lý)
Làm tương tự với 2 trường hợp còn lại:
\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
ĐKXĐ: x ≠ 0; x ≠ 2
\(< =>\dfrac{14x-28+20-4x}{16x\left(x-2\right)}=\dfrac{8x-8+2x}{16x\left(x-2\right)}\)
Suy ra: 14x - 28 + 20 - 4x = 8x - 8 + 2x
<=> 14x - 8x - 2x - 4x = 28 - 20 - 8
<=> 0x = 0
Vậy: S = { x | x ≠ 0;2 }
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
ĐKXĐ: x∉{0;2}
Ta có: \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(\Leftrightarrow\frac{5-x}{4x\left(x-2\right)}+\frac{7}{8x}-\frac{x-1}{2x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}=0\)
Suy ra: \(10-2x+7x-14-4x+4-x=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2\right\}\end{matrix}\right.\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
1. ĐKXĐ: $x\neq 1$
Sửa lại đề 1 chút:
$\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}$
$\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}-\frac{3x^2}{(x-1)(x^2+x+1)}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}$
$\Leftrightarrow x^2+x+1-3x^2=2x(x-1)$
$\Leftrightarrow 4x^2-3x-1=0$
$\Leftrightarrow (4x+1)(x-1)=0$
Vì $x\neq 1$ nên $x=-\frac{1}{4}$
2. ĐKXĐ: $x\neq 0;2$
PT \(\Leftrightarrow \frac{7}{8x}+\frac{5-x}{4x(x-2)}=\frac{x-1}{2x(x-2)}+\frac{1}{8(x-2)}\)
\(\Leftrightarrow \frac{7(x-2)}{8x(x-2)}+\frac{2(5-x)}{8x(x-2)}=\frac{4(x-1)}{8x(x-2)}+\frac{x}{8x(x-2)}\)
\(\Leftrightarrow 7(x-2)+2(5-x)=4(x-1)+x\)
\(\Leftrightarrow 5x-4=5x-4\) (luôn đúng)
Vậy pt có nghiệm $x\in\mathbb{R}$ với $x\neq 0;2$
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )
\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)
\(\Rightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow-3x+2x+x=10+4-14\)
\(\Leftrightarrow0=0\)
Vậy pt đã cho có nghiệm đúng với mọi x
\(\Leftrightarrow\) \(\dfrac{7}{8x}\)+\(\dfrac{5-x}{4x\left(x-2\right)}\)= \(\dfrac{x-1}{2x\left(x-2\right)}\)+ \(\dfrac{1}{8\left(x-2\right)}\)
\(\Rightarrow\) 7(x-2) + 2(5-x) = 4(x-1) +x
\(\Leftrightarrow\) 7x-2x+10-2x= 4x-4+x
\(\Leftrightarrow\)7x-2x-2x-4x-x = -4-10
\(\Leftrightarrow\) -2x = -14
\(\Leftrightarrow\) x = 7
Vậy phương trình có nghiệm x=7
⇔ 78x78x+5−x4x(x−2)5−x4x(x−2)= x−12x(x−2)x−12x(x−2)+ 18(x−2)18(x−2)
⇔ 7(x-2)8x(x-2)78x+2(5−x)8x(x−2)5−x4x(x−2)= 4(x−1)28x(x−2)x−12x(x−2)+ x8x(x−2)
18(x−2)
⇒7(x-2)+2(5-x)=4(x-1)+x
⇔ 7x-2x+10-2x= 4x-4+x
⇔7x-2x-2x-4x-x = -4-10
⇔ -2x = -14
⇔ x = 7
vậy tập của phương trình là: S=7}