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5 tháng 8 2018

a) \(3x\left(x-2\right)-4x+8=0\)

\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)

b) \(3\left(2x-1\right)^2+2-4x=0\)

\(\Leftrightarrow3\left(2x-1\right)^2-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(6x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\6x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{6}\end{cases}}\)

2 tháng 8 2018

\(a,3x\left(x-4\right)-2x+8=0\)

\(\Rightarrow3x\left(x-4\right)-2\left(x-4\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\4\end{cases}}}\)

Vậy \(x=\frac{2}{3}\)hoặc \(x=4\)

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

\(\Rightarrow9x^2-6x+1-\left(9x^2+3x+6x+1\right)-2=0\)

\(\Rightarrow9x^2-6x+1-9x^2-3x-6x-1-2=0\)

\(\Rightarrow-15x-2=0\)

\(\Rightarrow-15x=2\)

\(\Rightarrow x=\frac{-2}{15}\)

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: \(\left(3x-1\right)^2-9=0\)

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

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

30 tháng 6 2021

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

30 tháng 6 2021

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

30 tháng 7 2021

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)2 =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

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

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)2=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x2-x+3x-3=0

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

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

 

14 tháng 12 2021

\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

22 tháng 10 2023

\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(---\)

\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(---\)

\(c,4x(x-2)-x(3+4x)(?)\)

\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)

\(---\)

\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

\(---\)

\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(Toru\)

27 tháng 6 2021

\(a,=3x-9-4x+12=-x+3=0\)

\(\Leftrightarrow x=3\)

Vậy ..

\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy ..

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

\(\Leftrightarrow x=1\)

Vậy ..

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

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

Vậy ..

\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

Vậy ...

a) Ta có: 3(x-3)-4x+12=0

\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

hay x=3

Vậy: S={3}

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

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

\(\Leftrightarrow4x=-8\)

hay x=-2

Vậy: S={-2}

c) Ta có: \(x^3+3x=3x^2+1\)

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

\(\Leftrightarrow x-1=0\)

hay x=1

Vậy: S={1}

d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: S={0;2;-2}

 

19 tháng 2 2021

a)(3x-1)(4x-8)=0

⇔3x-1=0 hoặc 4x-8=0

1.3x-1=0⇔3x=1⇔x=1/3

2.4x-8=0⇔4x=8⇔x=2

phương trình có 2 nghiệm:x=1/3 và x=2

b)(x-2)(1-3x)=0

⇔x-2=0 hoặc 1-3x=0

1.x-2=0⇔x=2

2.1-3x=0⇔-3x=1⇔x=-1/3

phương trình có 2 nghiệm:x=2 và x=-1/3

c)(x-3)(x+4)-(x-3)(2x-1)=0

⇔(x+4)(2x-1)=0

⇔x+4=0 hoặc 2x-1=0

1.x+4=0⇔x=-4

2.2x-1=0⇔2x=1⇔x=1/2

phương trình có hai nghiệm:x=-4 và x=1/2

d)(x+1)(x+2)=2x(x+2)

⇔(x+1)(x+2)-2x(x+2)=0

⇔2x(x+1)=0

⇔2x=0 hoặc x+1=0

1.2x=0⇔x=0

2.x+1=0⇔x=-1

phương trình có 2 nghiệm:x=0 và x=-1