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14 tháng 12 2021

\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

23 tháng 10 2021

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

29 tháng 7 2021

`a)4x(x-2)+x-2=0`

`<=>(x-2)(4x+1)=0`

`<=>[(x-2=0),(4x+1=0):}`

`<=>[(x=2),(x=-1/4):}`

Vậy `S={2;-1/4}.`

`b)(3x-1)^3-9=0`

`<=>(3x-1-3)(3x-1+3)=0`

`<=>(3x-4)(3x+2)=0`

`<=>[(3x-4=0),(3x+2=0):}`

`<=>[(x=4/3),(x=-2/3):}`

Vậy `S={4/3;-2/3}.`

`c)x^3-8+(x-2)(x+1)=0`

`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`

`<=>(x-2)(x^2+3x+5)=0`

Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`

`<=>x-2=0`

`<=>x=2`

Vậy `S={2}`

a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)

b)Ta có: \(\left(3x-1\right)^2-9=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

6 tháng 8 2021

a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)

b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)

c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)

a) Ta có: \(4x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)

b) Ta có: \(\left(3x-1\right)^2-9=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

c: =>(x-1)(x+1)=0

hay \(x\in\left\{1;-1\right\}\)

2 tháng 1 2022

plss

21 tháng 10 2023

a) \(\left(2x+1\right)\left(x-2\right)-2x^2=0\)

\(\Leftrightarrow2x^2-4x+x-2-2x^2=0\)

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

\(\Leftrightarrow-3x-2=0\)

\(\Leftrightarrow-3x=2\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

b) \(\left(x+3\right)\left(2x-1\right)+x^2=9\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+x^2-9=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{4}{3}\end{matrix}\right.\)

`#3107.101107`

a)

`(2x + 1)(x - 2) - 2x^2 = 0`

`<=> 2x^2 - 3x - 2 - 2x^2 = 0`

`<=> -3x - 2 = 0`

`<=> -3x = 2`

`<=> x = -2/3`

Vậy, `x=-2/3`

b)

`(x + 3)(2x - 1) + x^2 = 9`

`<=> 2x^2 - 5x - 3 + x^2 = 9`

`<=> 3x^2 - 5x - 3 = 9`

`<=> 3x^2 - 3x - 12 = 0`

`<=> 3x^2 + 4x - 9x - 12 = 0`

`<=> (3x^2 - 9x) + (4x - 12) = 0`

`<=> 3x(x - 3) + 4(x - 3) = 0`

`<=> (3x + 4)(x - 3) = 0`

`<=>` TH1: `3x + 4 = 0`

`<=> 3x = -4`

`<=> x = -4/3`

TH2: `x - 3 = 0`

`<=> x = 3`

Vậy,` x \in {-4/3; 3}.`

a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)

\(\Leftrightarrow x^3-27-x^3+3x=0\)

\(\Leftrightarrow x=9\)

b: Ta có: \(8x^4+x=0\)

\(\Leftrightarrow x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

30 tháng 6 2021

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)