K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 7 2021

\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
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}

30 tháng 4 2017

\(\Leftrightarrow\) \(\dfrac{7}{8x}\)+\(\dfrac{5-x}{4x\left(x-2\right)}\)= \(\dfrac{x-1}{2x\left(x-2\right)}\)+ \(\dfrac{1}{8\left(x-2\right)}\)

\(\Rightarrow\) 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

ok

11 tháng 2 2019

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}

17 tháng 5 2017

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:

begin{align} \begin{cases} -2 bé hơn hoặc bằng x bé hơn -1 \\ x lớn hơn hoặc bằng -1 \\ \end{cases} \end{align}
26 tháng 5 2021

 

\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( 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

26 tháng 5 2021

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

a) Ta có: \(\dfrac{1}{x+3}+\dfrac{8-x}{4x^2+8x}\)

\(=\dfrac{1}{x+3}+\dfrac{8-x}{4x\left(x+2\right)}\)

\(=\dfrac{4x\left(x+2\right)}{4x\left(x+3\right)\left(x+2\right)}+\dfrac{\left(8-x\right)\left(x+3\right)}{4x\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+8x+8x+24-x^2-3x}{4x\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{3x^2+13x+24}{4x\left(x+3\right)\left(x+2\right)}\)

b) Ta có: \(\dfrac{3-2x}{\left(x-5\right)\left(x+2\right)}+\dfrac{1}{x+5}\)

\(=\dfrac{\left(3-2x\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)\left(x+2\right)}+\dfrac{\left(x-5\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{3x+15-2x^2-10x+x^2+2x-5x-10}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{-x^2-10x+5}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}\)