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24 tháng 7 2016

        (2x - 1)(4x2 + 2x + 1) = x(x-8)

<=>                   (2x)3 - 1= x2 - 8x

<=>     (8x3 - x2) + (8x - 1) = 0

<=>     x2(8x - 1) + (8x - 1) = 0Σ

<=>            (x2 + 1)(8x - 1) = 0

  • x2 + 1 = 0 => x2 = -1 ( vô lý )
  • 8x - 1 = 0 => 8x = 1 => x = 0,125  

 

 

 

 

1 tháng 9 2020

\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)

\(\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

     \(\text{Vậy x=-5}\)

\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)

\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)

\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)

\(\Rightarrow-16x-8=7\)

\(\Rightarrow-16x=15\)

\(\Rightarrow x=\frac{-15}{16}\)

      \(\text{Vậy }x=\frac{-15}{16}\)

\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)

\(\Rightarrow-9+8x-1=8\)

\(\Rightarrow8x=18\)

\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

      \(\text{Vậy }x=\frac{9}{4}\)

\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)

14 tháng 10 2021

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

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

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

14 tháng 9 2021

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)

28 tháng 5 2022

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

28 tháng 5 2022

`A`

8 tháng 9 2021

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

20 tháng 7 2018

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x2 - 10x - 2x2 + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5

<=> 2x+ x + 6x + 3 - 2x2 - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

19 tháng 7 2018

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

2 tháng 9 2021

a) \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Rightarrow3x-6+2x-6=5\)

\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)

b) \(\left(2x-8\right)^2-16=0\)

\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)

\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)

a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)