Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé 

a, \(\left|5x\right|=x+2\)

Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)

Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)

b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)

Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )

Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )

Vậy phương trình vô nghiệm 

a) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)

Suy ra: \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)

a/ \(\dfrac{3x^2+7x-10}{x}=0\)

\(< =>3x^2+7x-10=0\)

\(< =>3x^2+10x-3x-10=0\)

\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)

\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)

\(< =>\left(3x+10\right)\left(x-1\right)=0\)

\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)

Vậy tập nghiệm của .....

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

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

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

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

Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)

b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)

\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)

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

Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)

Đây là toán 9 chứ

a) Ta có: \(2x^3+5x^2-3x=0\)

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

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

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

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

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

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

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

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

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

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!

\(\left|3x^2-7x+2\right|=-x^2+5x-6\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-7x+2=-x^2+5x-6\\-\left(3x^2-7x+2\right)=-x^2+5x-6\end{matrix}\right.\)

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

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

vậy....

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

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

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

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

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

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

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

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

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

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

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

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

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

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

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

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

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

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

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

Vậy \(S=\left\{3;-2\right\}\)

Chúc bạn học tốt !!!

\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

\(\Leftrightarrow\frac{x^2+2x-3x-6}{x-3}=0\)

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

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

<=> x + 2 = 0

=> x = -2

(x² + x  + 1)(6 - 2x) = 0

<=> 6 - 2x = 0 (do x² + x + 1 > 0)

<=> 2x = 6

<=> x = 3

Vậy S = {3}

(8x - 4)(x² + 2x + 2) = 0

<=> 8x - 4 = 0 (vì x² + 2x + 2 > 0)

<=> 8x = 4

<=> x = 1/2 

Vậy S  = {1/2}

x³ - 7x + 6 = 0

<=> x³ - x - 6x + 6 = 0

<=> x(x² - 1) - 6(x - 1) = 0

<=> x(x - 1)(x + 1) - 6(x - 1) = 0

<=> (x² + x - 6)(x - 1) = 0

<=> (x² + 3x - 2x - 6)(x - 1) = 0

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

<=> x + 3 = 0

hoặc x - 2 = 0

hoặc x  - 1 = 0

<=> x = -3

hoặc x = 2

hoặc x = 1

Vậy S = {-3; 1; 2}

x⁵ - 5x³ + 4x = 0

<=> x(x⁴ - 5x² + 4) = 0

<=> x(x⁴ - x² - 4x² + 4) = 0

<=> x[x²(x² - 1) - 4(x² - 1)] = 0

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

<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x  + 1 = 0

<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1

Vậy S = {-2; -1; 0; 1; 2}

+ Ta có: \(\left(x^2+x+1\right).\left(6-2x\right)=0\)

 - Ta lại có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

- Vì \(x^2+x+1>0\forall x\)mà \(\left(x^2+x+1\right).\left(6-2x\right)=0\)

  \(\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)\)

Vậy \(S=\left\{3\right\}\)

+ Ta có: \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)

 - Ta lại có: \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\)

 - Vì \(x^2+2x+2>0\forall x\)mà \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)

   \(\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)

+ Ta có: \(x^3-7x+6=0\)

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

       \(\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0\)

       \(\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0\) 

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

 Vậy \(S=\left\{-3;1;2\right\}\)

 + Ta có: \(x^5-5x^3+4x=0\)

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

       \(\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0\)

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

       \(\Leftrightarrow x=0\left(TM\right)\)

hoặc  \(x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)\)

hoặc \(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)\)

Vậy \(S=\left\{-2;-1;0;1;2\right\}\)

!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!