a: =>x(x+3)=0

=>x=0 hoặc x=-3

b: =>x(1-2x)=0

=>x=0 hoặc x=1/2

c: =>(x-7)(2x+3-x)=0

=>(x-7)(x+3)=0

=>x=7 hoặc x=-3

d: =>(x-2)(3x-1-x-3)=0

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

=>x=2

a)

`x^2 +3x=0`

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

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

b)

`x-2x^2 =0`

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

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

c)

`(x-7)(2x+3)=x(x-7)`

`<=>(x-7)(2x+3)-x(x-7)=0`

`<=>(x-7)(2x+3-x)=0`

`<=>(x-7)(x+3)=0`

\(< =>\left[{}\begin{matrix}x-7=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

d)

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

`<=>(x-2)(x+3)-(x-2)(3x-1)=0`

`<=>(x-2)(x+3-3x+1)=0`

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

\(< =>\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\\ < =>x=2\)

a)\(x^2+3x=0\)

<=>x(x+3)=0

x=0 hoặc x+3=0

x=0 hoặc x=-3

b)x-2x²=0

x(1-2x)=0

x=0 hoặc 1-2x=0

x=0 hoặc x=0,5

c)(x-7)(2x+3)=4(x-7)

(x-7)(2x+3)-4(x-7)=0

(x-7)(2x+3-4)=0

x-7=0 hoặc 2x+3-4=0

x=7 hoặc x=0,5

d)(x-2)(x+3)=(x-2)(3x-1)

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

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

x-2=0 hoặc x+3-3x+1=0

x=2 hoặc x-3x=-3-1

               -2x=-4

               x=2

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

Đề bài không rõ yêu cầu. Bạn coi lại đề.

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

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

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

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

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

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

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

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

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

=>\(\dfrac{3x^3-9x^2+9x-2x^3+2x^2-6x}{\left(x^2-3x+3\right)\left(x^2-x+3\right)}=-1\)

=>x^3-7x^2+3x=-[(x^2+3)^2-4x(x^2+3)+3x^2]

=>x^3-7x^2+3x+(x^2+3)^2-4x(x^2+3)+3x^2=0

=>x^3-4x^2+3x+x^4+6x^2+9-4x^3-12x=0

=>x^4-3x^3+2x^2-9x+9=0

=>(x-3)(x-1)(x^2+x+3)=0

=>x=3;x=1

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

Ta có : \(x^2-x+3=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)

\(x^2-3x+3=x^2-3x+\dfrac{9}{4}+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\)

\(\RightarrowĐKXĐ:x\in R\)

Đặt : \(t=x^2-x+3\)

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

\(\Leftrightarrow3x\left(t-2x\right)-2xt+t\left(t-2x\right)=0\)

\(\Leftrightarrow t^2-xt-6x^2=0\)

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

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

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

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

Thay ​\(t=x^2-x+3\) ​lại vào (b) được :

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

Mà : \(x^2-4x+3=x^2-x-3x+3\)

\(=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\left(c'\right)\)

và : \(x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\left(c''\right)\)

Thay (c') và (c'') vào (c) được :

\(\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-1=0\Leftrightarrow x=1\left(tmđk\right)\\x-3=0\Leftrightarrow x=3\left(tmđk\right)\end{matrix}\right.\\\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\Leftrightarrow x\in\varnothing\end{matrix}\right.\)

Vậy : Phương trình có tập nghiệm \(S=\left\{1;3\right\}\)

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

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

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

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

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

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

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

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

a) |3x| = x + 6 (1)

Ta có 3x = 3x khi x ≥ 0 và 3x = -3x khi x < 0

Vậy để giải phương trình (1) ta quy về giải hai phương trình sau:

+ ) Phương trình 3x = x + 6 với điều kiện x ≥ 0

Ta có: 3x = x + 6 ⇔ 2x = 6 ⇔ x = 3 (TMĐK)

Do đó x = 3 là nghiệm của phương trình (1).

+ ) Phương trình -3x = x + 6 với điều kiện x < 0

Ta có -3x = x + 6 ⇔ -4x + 6 ⇔ x = -3/2 (TMĐK)

Do đó x = -3/2 là nghiệm của phương trình (1).

Vậy tập nghiệm của phương trình đã cho S = {3; -3/2}

ĐKXĐ: x ≠ 0, x ≠ 2

Quy đồng mẫu hai vễ của phương trình, ta được:

Vậy tập nghiệm của phương trình là S = {-1}

c) (x + 1)(2x – 2) – 3 > –5x – (2x + 1)(3 – x)

⇔ 2x2 – 2x + 2x – 2 – 3 > –5x – (6x – 2x2 + 3 – x)

⇔ 2x2 – 5 ≥ –5x – 6x + 2x2 – 3 + x

⇔ 10x ≥ 2 ⇔ x ≥ 1/5

Tập nghiệm: S = {x | x ≥ 1/5}