1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3) 
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8 
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5 
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27 
<=> - 3x^2 - 18x - 22 = 0 
<=> 3x^2 + 18x + 22 = 0 

Nửa chu vi mảnh đất là: 

                                               120 : 2 = 60 (m)

Chiều dài hơn chiều rộng là:

                                               5 + 5 = 10 (m)

Chiều rộng là:

                                          ( 60 - 10 ) : 2 = 25 (m)

Chiều dài là:

                                                25 + 10 = 35 (m)

Diện tích là:

                                               25  35 = 875 ( )

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8+3\left(x^2-6x+9\right)=x^3+6x^2+12x+8-x^3-9x+3x^2+27-6x^2+5\)

\(\Leftrightarrow-3x^2+3x-9+3\left(x^2-6x+9\right)=3x^2+3x+40\)

\(\Leftrightarrow-3x^2+3x-9+3x^2-18x+27-3x^2-3x-40=0\)

\(\Leftrightarrow-3x^2-18x-22=0\)

\(\Leftrightarrow3x^2+18x+22=0\)

\(\Delta=18^2-4\cdot3\cdot22=60\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-18-2\sqrt{15}}{6}=\dfrac{-9-\sqrt{15}}{3}\\x_2=\dfrac{-18+2\sqrt{15}}{6}=\dfrac{-9+\sqrt{15}}{3}\end{matrix}\right.\)

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

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

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

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

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)

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

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

`Answer:`

a. \(x^3+6x^2+12=19\)

\(\Leftrightarrow x^3+6x^2+12x-19=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)

\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)

Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)

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

b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)

\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)

\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)

\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)

c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)

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

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

\(=\left(x+5\right)^2\)

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

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

\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)

\(\left(3x+1\right)^2=\left(\pm2\right)^2\)

\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)

\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)

\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)

***

\(\left(x+3\right)\left(3-x\right)=5\)

\(3^2-x^2=5\)

\(x^2=9-5\)

\(x^2=4\)

\(x^2=\left(\pm2\right)^2\)

\(x=\pm2\)

***

\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)

\(27x^3+3=2\)

\(27x^3=2-3\)

\(\left(3x\right)^3=-1\)

\(3x=-1\)

\(x=-\frac{1}{3}\)

Đâu có y đâu bạn

số điện thoại của bạn là gì ?

Ít thôi -..-

a) ( 3x + 2 )( 2x + 9 )  - ( x + 3 )( 6x + 1 ) = ( x + 1 )² - ( x + 2 )( x - 2 )

<=> 6x² + 31x + 18 - ( 6x² + 19x + 3 ) = x² + 2x + 1 - ( x² - 4 )

<=> 6x² + 31x + 18 - 6x² - 19x - 3 = x² + 2x + 1 - x² + 4

<=> 12x + 15 = 2x + 5

<=> 12x - 2x = 5 - 15

<=> 10x = -10

<=> x = -1

b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )

<=> 2x² - 5x - 12 + x² - 7x + 10 = 3x² - 17x + 20

<=> 3x² - 12x - 2 = 3x² - 17x + 20

<=> 3x² - 12x - 3x² + 17x = 20 + 2

<=> 5x = 22

<=> x = 22/5

c) ( x + 2 )³ - ( x - 2 )³ - 12x( x - 1 ) = -8

<=> x³ + 6x² + 12x + 8 - ( x³ - 6x² + 12x - 8 ) - 12x² + 12x = -8

<=>  x³ + 6x² + 12x + 8 - x³ + 6x² - 12x + 8 - 12x² + 12x = -8

<=> 12x + 16 = -8

<=> 12x = -24

<=> x = -2

d) ( 3x - 1 )² - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )² = 16

<=> 9x² - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x² - 2x + 1 ) = 16

<=> 9x² - 11x - 3 - x² + 2x - 1 = 16

<=> 8x² - 9x - 4 = 16

<=> 8x² - 9x - 4 - 16 = 0

<=> 8x² - 9x - 20 = 0

( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm

                                                         2 là nghiệm vô tỉ =) )

a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)² - (x + 2)(x - 2)

=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x² + 2x + 1 - x(x - 2) - 2(x - 2)

=> 6x² + 27x + 4x + 18 - 6x² - x - 18x - 3 = x² + 2x + 1 - x² + 2x - 2x + 4

=> (6x² - 6x²) + (27x + 4x - x - 18x) + (18 - 3) = (x² - x²) + (2x + 2x - 2x) + (1 + 4)

=> 12x + 15 = 2x + 5

=> 12x + 15  - 2x - 5 = 0

=> 10x + 10 = 0

=> 10x = -10 => x = -1

b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)

=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20

=> (2x² + x²) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x² - 17x + 20

=> 3x² - 12x - 2 = 3x² - 17x + 20

=> 3x² - 12x - 2 - 3x² + 17x - 20 = 0

=> (3x² - 3x²) + (-12x + 17x) + (-2 - 20) = 0

=> 5x - 22 = 0

=> 5x = 22 => x = 22/5

c) (x + 2)³ - (x - 2)³ - 12x(x - 1) = -8

=> x³ + 6x² + 12x + 8 - (x³  - 6x² + 12x - 8) - 12x² + 12x = -8

=> x³ + 6x² + 12x + 8 -x³ + 6x² - 12x + 8 - 12x² + 12x = -8

=> (x³ - x³) + (6x² + 6x² - 12x²) + (12x - 12x + 12x) + (8 + 8) = -8

=> 12x + 16 = -8

=> 12x = -24

=> x = -2

Còn bài cuối làm nốt