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

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

(x + 1)(x + 2) = (2 - x)(x + 2)

<=> x² + 2x + x + 2 = 4 - x²

<=> x² + 3x + 2 = 4 - x²

<=> x² + 3x + 2 - 4 + x² = 0

<=> 2x² + 3x - 2 = 0

<=> 2x² + 4x - x - 2 = 0

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

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

<=> x + 2 = 0 hoặc 2x - 1 = 0

<=> x = -2 hoặc x = 1/2

còn bài 2 nữa bạn

ư ư :<

a) 2(m + 3/5) - (m + 13/5) = 5

<=> 2m + 6/5 - m - 13/5 = 5

<=> m - 7/5 = 5

<=> m = 5 + 7/5

<=> m = 32/5

b) \(\frac{2\left(3m+1\right)+1}{5}-\frac{2\left(3m-1\right)}{5}+\frac{3m+1}{10}=5\)

<=> 15(2m + 1) - 8(3m - 1) + 3(3m + 2) = 100

<=> 30m + 15 - 24m + 1 + 9m + 6 = 100

<=> 12m + 27 = 100

<=> 12m = 100 - 27

<=> 12m = 73

<=> m = 73/12

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

lấy cả 2 vế trừ cho nhau = 0,từ đó tính ra x

bạn ơi không ra được đâu

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 .

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)

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 .