Bài 2:

(1 + x)³ + (1 - x)³ - 6x(x + 1) = 6

<=> x³ + 3x² + 3x + 1 - x³ + 3x² - 3x + 1 - 6x² - 6x = 6

<=> -6x + 2 = 6

<=> -6x = 6 - 2

<=> -6x = 4

<=> x = -4/6 = -2/3

Bài 3: 

a) (7x - 2x)(2x - 1)(x + 3) = 0

<=> 10x³ + 25x² - 15x = 0

<=> 5x(2x - 1)(x + 3) = 0

<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0

<=> x = 0 hoặc x = 1/2 hoặc x = -3

b) (4x - 1)(x - 3) - (x - 3)(5x + 2) = 0

<=> 4x² - 13x + 3 - 5x² + 13x + 6 = 0

<=> -x² + 9 = 0

<=> -x² = -9

<=> x² = 9

<=> x = +-3

c) (x + 4)(5x + 9) - x² + 16 = 0

<=> 5x² + 9x + 20x + 36 - x² + 16 = 0

<=> 4x² + 29x + 52 = 0

<=> 4x² + 13x + 16x + 52 = 0

<=> 4x(x + 4) + 13(x + 4) = 0

<=> (4x + 13)(x + 4) = 0

<=> 4x + 13 = 0 hoặc x + 4 = 0

<=> x = -13/4 hoặc x = -4

Lê Nhật Hằng cảm ơn bạn nha

T a   c ó :   A = B   ⇔   ( x + 2 ) ( x – 2 ) + 3 x 2 = ( 2 x + 1 ) 2 + 2 x   ⇔   x 2 – 4 + 3 x 2 = 4 x 2 + 4 x + 1 + 2 x ⇔   x 2 + 3 x 2 – 4 x 2 – 4 x – 2 x = 1 + 4   ⇔ - 6 x = 5 ⇔   x   =   -   5 / 6     V ậ y   v ớ i   x = -   5 / 6   t h ì   A   =   B .

T a   c ó :   A = B   ⇔   ( x – 3 ) ( x + 4 ) – 2 ( 3 x – 2 ) = ( x – 4 ) 2   ⇔   x 2 + 4 x – 3 x – 12 – 6 x + 4 = x 2 – 8 x + 16 ⇔   x 2 – x 2 + 4 x – 3 x – 6 x + 8 x = 16 + 12 – 4 ⇔   3 x = 24 ⇔   x   =   8     V ậ y   v ớ i   x   =   8   t h ì   A   =   B

T a   c ó :   A   =   B   ⇔   ( x + 1 ) 3 – ( x – 2 ) 3 = ( 3 x – 1 ) ( 3 x + 1 ) ⇔   x 3 + 3 x 2 + 3 x + 1 – x 3 + 6 x 2 – 12 x + 8 = 9 x 2 – 1 ⇔   x 3 – x 3 + 3 x 2 + 6 x 2 – 9 x 2 + 3 x – 12 x = - 1 – 1 – 8 ⇔   - 9 x = - 10   ⇔   x   =   10 / 9     V ậ y   v ớ i   x   =   10 / 9   t h ì   A   =   B .

T a   c ó :   A   =   B   ⇔   ( x – 1 ) ( x 2 + x + 1 ) – 2 x = x ( x – 1 ) ( x + 1 ) ⇔   x 3 – 1 – 2 x = x ( x 2 – 1 )   ⇔   x 3 – 1 – 2 x = x 3 – x   ⇔   x 3 – x 3 – 2 x + x = 1 ⇔ - x = 1 ⇔ x = - 1     V ậ y   v ớ i   x = - 1   t h ì   A   =   B

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

\(a,\dfrac{y-1}{y-2}-\dfrac{y+3}{y-4}=\dfrac{-2}{\left(y-2\right)\left(y-4\right)}\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y-4\right)-\left(y+3\right)\left(y-2\right)+2}{\left(y-2\right)\left(y-4\right)}=0\)\(\left(dkxd:y\ne4;2\right)\)

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

\(\Leftrightarrow-6y+12=0\)

\(\Leftrightarrow y=2\)\(\left(ktm\right)\)

Vậy ko có bất kì giá trị y nào để 2 biểu thức bằng nhau

\(b,\dfrac{8y}{y-7}+\dfrac{1}{7-y}=8\)

\(\Leftrightarrow\dfrac{8y}{y-7}-\dfrac{1}{y-7}=8\)\(\left(dkxd:y\ne7\right)\)

\(\Leftrightarrow8y-1-8\left(y-7\right)=0\)

\(\Leftrightarrow8y-1-8y+56=0\)(Vô lý)

Vậy ko có bất kì giá trị y nào để biểu thức có giá trị = 8