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

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

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

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

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

Vậy : ...

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

\(\Leftrightarrow9x^2-24x+16-4\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow9x^2-24x+16-4x^2-8x-4=0\)

\(\Leftrightarrow5x^2-32x+12=0\)

\(\Leftrightarrow5x^2-30x-2x+12=0\)

\(\Leftrightarrow5x\left(x-6\right)-2\left(x-6\right)=0\)

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

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

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

Vậy tập nghiệm của phương trình là : \(S=\left\{6;\frac{2}{5}\right\}\)

2/ \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4+2x^3-3x^2-6x-2x-4=0\)

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

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

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

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

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

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

hoặc   \(x+1=0\)

hoặc   \(x-2=0\)

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

hoặc   \(x=-1\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-1\right\}\)

3/ \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)-72=0\)

Đặt \(t=x^2-4\), ta có :

\(t\left(t-6\right)-72=0\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow t^2-12t+6t-72=0\)

\(\Leftrightarrow t\left(t-12\right)+6\left(t-12\right)=0\)

\(\Leftrightarrow\left(t+6\right)\left(t-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+6=0\\t-12=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2=0\left(ktm\right)\\x^2-16=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x=\pm4\)

Vậy tập nghiệm của phương trình là : \(S=\left\{4;-4\right\}\)

4/ \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

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

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

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

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc   \(x+2=0\)

hoặc  \(2x+1=0\)

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

hoặc  \(x=-2\)

hoặc  \(x=-\frac{1}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-2;\frac{1}{2}\right\}\)

(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 ^_^#@@!!      

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

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

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

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

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

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

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

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

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

`<=>2x=4<=>x=2`

len google di ban

mk chua hoc bai nay

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 .....

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

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

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

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

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

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

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

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

`12)x(x-3)+3(x-3)=0`

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

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

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

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

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)