\(a,\Rightarrow4x^2-1-4x^2+2x=5\\ \Rightarrow2x=6\Rightarrow x=3\\ b,\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-4\right)=0\\ \Rightarrow\left(x+1\right)\left(x+2\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=2\end{matrix}\right.\)

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Khó quá,hay quá !!!

CHÚC EM HỌC TỐT NHÉ 

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

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

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

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a: =>(x-3)(x+1)=0

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

c: =>(x-5)(x+1)=0

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

h: =>x^2-4x+4-3=0

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

Bài 1. a) 4x - 3 = 0

⇔ x = \(\dfrac{3}{4}\)

KL.....

b) - x + 2 = 6

⇔ x = - 4

KL...

c) -5 + 4x = 10

⇔ 4x = 15

⇔ x = \(\dfrac{15}{4}\)

KL....

d) 4x - 5 = 6

⇔ 4x = 11

⇔ x = \(\dfrac{11}{4}\)

KL....

h) 1 - 2x = 3

⇔ -2x = 2

⇔ x = -1

KL...

Bài 2. a) ( x - 2)( 4 + 3x ) = 0

⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)

KL......

b) ( 4x - 1)3x = 0

⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)

KL.....

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

⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)

KL.....

d) 3x( x + 2) = 0

⇔ x = 0 hoặc x = -2

KL.....

Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0

⇔ x - 10 ≥ 0

⇔ x ≥ 10

b) 3 - 2( 2x + 3) ≤ 9x - 4

⇔ - 4x - 3 ≤ 9x - 4

⇔ 13x ≥1

⇔ x ≥ \(\dfrac{1}{13}\)

(4x - 1)(3 - x) - (2x + 3)(2x - 1) = 0

12x + \(4x^2\)- 3 + x - 4x + 2x - 6x + 3 = 0 

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

4x . 4x + 5x                                           = 0

  16x    + 5x                                           = 0

       21x                                                 = 0

           x                                                 = 0

2x(8x + 3) - (4x - 1)(4x + 1) = 13

16x + 6x   - 16x - 4x + 4x +1 = 13

6x +1                                     = 13

6x                                          = 12

  x                                           = 2

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

\(n,3x^3-3x^2-6x=0\)

\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)

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

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)

\(m,x^2-11x+28=0\)

\(\Leftrightarrow x^2-4x-7x+28=0\)

\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)

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

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

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

\(\Leftrightarrow6x=6\)

hay x=1

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

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

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

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

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

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

<=> (x - 1)(x² + x + 1) - x(x - 1) = 0

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

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

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

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

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

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

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

\(3x.x-3x+2x-2=0\)

\(2x-2=0\)

\(2x=2\)

\(x=1\)