a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)

\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+5x\left(x-2\right)+2\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+5x+2\right)=0\)

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

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

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

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

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

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

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

\(\Rightarrow x\in\left\{2;-1;-2\right\}\)

Vậy....

c, \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[2\left(x^2-x+1\right)+7x\right]=0\)

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

Tập nghiệm của pt: \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)

b, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\) (1)

Đặt: \(x^2-7=t\left(t\ge-7\right)\)

Khi đó (1) trở thành: \(\left(t+3\right)\left(t-3\right)=72\Leftrightarrow t^2-9=72\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\left(loai\right)\end{cases}}\)

\(t=9\Rightarrow x^2-7=9\Leftrightarrow x=\pm4\)

Tập nghiệm của pt là \(S=\left\{\pm4\right\}\)

a, \(x^4+2x^3-3x^2-8x-4=0\)

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

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

\(\Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\) ( 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 \(S=\left\{-1;-2;\frac{1}{2}\right\}\)

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 .....

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: =>(x-1)(3x-4)>0

=>x>4/3 hoặc x<1

b: =>x^3-3x^2-10x^2+30x+12x-36>0

=>(x-3)(x^2-10x+12)>0

Th1: x-3>0và x^2-10x+12>0

=>x>5+căn 13

TH2: x-3<0 và x^2-10x+12<0

=>x<3 và 5-căn 13<x<5+căn 13

=>3<x<5+căn 13

a) (3x² - 7x – 10)[2x² + (1 - √5)x + √5 – 3] = 0

=> hoặc (3x² - 7x – 10) = 0 (1)

hoặc 2x² + (1 - √5)x + √5 – 3 = 0 (2)

Giải (1): phương trình a - b + c = 3 + 7 - 10 = 0

nên

x₁ = - 1, x₂ = =

Giải (2): phương trình có a + b + c = 2 + (1 - √5) + √5 - 3 = 0

nên

x₃ = 1, x₄ =

b) x³ + 3x²– 2x – 6 = 0 ⇔ x²(x + 3) – 2(x + 3) = 0 ⇔ (x + 3)(x² - 2) = 0

=> hoặc x + 3 = 0

hoặc x² - 2 = 0

Giải ra x₁ = -3, x₂ = -√2, x₃ = √2

c) (x² - 1)(0,6x + 1) = 0,6x² + x ⇔ (0,6x + 1)(x² – x – 1) = 0

=> hoặc 0,6x + 1 = 0 (1)

hoặc x² – x – 1 = 0 (2)

(1) ⇔ 0,6x + 1 = 0

⇔ x₂ = =

(2): ∆ = (-1)² – 4 . 1 . (-1) = 1 + 4 = 5, √∆ = √5

x₃ = , x₄ =

Vậy phương trình có ba nghiệm:

x₁ = , x₂ = , x₃ = ,

d) (x² + 2x – 5)² = ( x² – x + 5)² ⇔ (x² + 2x – 5)² - ( x² – x + 5)² = 0

⇔ (x² + 2x – 5 + x² – x + 5)( x² + 2x – 5 - x² + x - 5) = 0

⇔ (2x² + x)(3x – 10) = 0

⇔ x(2x + 1)(3x – 10) = 0

Hoặc x = 0, x = , x =

Vậy phương trình có 3 nghiệm:

x₁ = 0, x₂ = , x₃ =

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

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):}`