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22 tháng 9 2021

\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

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

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

9 tháng 12 2021

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

25 tháng 12 2022

\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)

\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)

Vậy \(x=\dfrac{-255}{2}\)

25 tháng 12 2022

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

a) Ta có: \(7x^2-28=0\)

\(\Leftrightarrow7\left(x^2-4\right)=0\)

\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)

mà 7>0

nên (x-2)(x+2)=0

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

Vậy: \(x\in\left\{2;-2\right\}\)

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

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

mà \(\dfrac{2}{3}>0\)

nên x(x-2)(x+2)=0

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

Vậy: \(x\in\left\{0;-2;2\right\}\)

c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)

d) Ta có: \(\left(2x-1\right)^2-25=0\)

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

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-2\right\}\)

11 tháng 1 2021

a,7x2 - 28 = 0

=> 7x2 = 28 => x2 = 4 => x = 2

b,2/3x(x2 - 4) = 0

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

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

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

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)2 - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

4 tháng 10 2021

\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

4 tháng 10 2021

a) \(\Rightarrow4x\left(x^2-9\right)=0\)

\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)

\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

12 tháng 10 2021

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

18 tháng 10 2021

ỳtct7ct7c7c7t79tc9

 

21 tháng 9 2021

a. 9x2 - 6x - 3 = 0

<=> 3(3x2 - 2x - 1) = 0

<=> 3(3x2 - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

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

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

b. (2x + 1)2 - 4(x + 2)2 = 9

<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

21 tháng 9 2021

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

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

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

6 tháng 11 2021

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)

6 tháng 11 2021

 

\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)

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

\(6x+18=0\)

\(6x=-18\)

\(x=-3\)

Vậy x=-3

\(b,5x^3+20x=0\)

\(5x\left(x^2+4\right)=0\)

\(Th1:5x=0=>x=0\)

\(Th2:x^2+4=0\)

\(x^2=-4\)(vô lý)

Vậy x=0

27 tháng 8 2021

\(a,x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow x^2+5x-x^2-x+6=0\Leftrightarrow4x=-6\\ \Leftrightarrow x=-\dfrac{3}{2}\)

\(b,2x^3-18x=0\\ \Leftrightarrow2x\left(x^2-9\right)=0\\ \Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

a: Ta có: \(x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+5x-x^2-3x+2x+6=0\)

\(\Leftrightarrow7x=-6\)

hay \(x=-\dfrac{6}{7}\)

b: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

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

10 tháng 9 2021

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

10 tháng 9 2021

a)5x.(x+1)-5.(x+1).(x-2)=0

⇒5x(x+1)-(5x-10)(x+1)=0

⇒(x+1)(5x-5x+10)=0

⇒10(x+1)=0

⇒x+1=0⇒x=-1