Simplification: (1) |3x-2 | + |2x + 3 |; (2) |x-1 | - 3 | + |3x + 1 |

Simplification: (1) |3x-2 | + |2x + 3 |; (2) |x-1 | - 3 | + |3x + 1 |

(1) When 3x-2 & lt; 0, 2x + 3 & lt; 0, namely X & lt; - 32, the original formula = 2-3x-2x-3 = - 5x-1; when 3x-2 ≥ 0, 2x + 3 ≥ 0, namely x ≥ 23, the original formula = 3x-2 + 2x + 3 = 5x + 1; when 3x-2 ≥ 0, 2x + 3 & lt; 0, namely - 32 ≤ X & lt; 23, the original formula = 2

Simplification: (1) |3x-2 | + |2x + 3 |; (2) |x-1 | - 3 | + |3x + 1 |

(1) When 3x-2 & lt; 0, 2x + 3 & lt; 0, i.e. X & lt; - 32, the original formula = 2-3x-2x-3 = - 5x-1; when 3x-2 ≥ 0, 2x + 3 ≥ 0, i.e. x ≥ 23, the original formula = 3x-2 + 2x + 3 = 5x + 1; when 3x-2 ≥ 0, 2x + 3 & lt; 0, X does not exist; 3x-2 & lt; 0, 2x + 3 ≥ 0, i.e. - 32 ≤ X & lt; The original formula = 2-3x + 2x + 2x + 3 = - x + 5; (2) when |x-1-1 |-1 | -1 | -1 | -3-3 = x-1-1-1-3-3 ≥ 0, 3x-3 = 2-3x + 2x + 2x + 3 = 2x-3; (2) when |x-1 |-1-3 ≥ 0, 3x + 1-3 ≥ 0, 3x-1-1 |-1 |-1 |-1 |-3 = 1-x-3-x-3-3-x-3 & gt; 0, this is the X & lt;;;; - 2 and X & gt; - 2 and the X & gt;; 13, the X & gt; \\\\\\ifx-1 | - 3 = x-1-3 ≥ 0, X & gt; 4 and X & lt; - 13, then x does not exist; ④ X-1 & lt; The original formula = - 4x-3; when |-x-1-1-1-3-3-3 & gt; 0, x-1-3-3 = 1-x-3 & gt; 0, 3x + 1 & lt; 0, 3x + 1 & lt; 0, 3x-1-3-3 = 1-x-3-3 & gt; 0, 6-x-1-1-3-3 & gt; 0, x-1-1-1-1-3-3-3-3-3-3-3-3-3 & lt; 0, when |-1-1-1-6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-3-3-x-x-3-3-x-x-3-3-3 ≤ 0, x-x-x-x-x-3-3-x-3 ≤ 0, x-x-3-3, x-3-when | X-1 | - 3 ≤ 0, 3x + 1 ≥ 0, and | X-1 | - 3 = x-1-3 ≤ 0, X & lt; The answer is: (1) - 5x-1 (X & lt; 4, the original formula = 2x + 5; 4, the original formula = 2x & lt; 4, the original formula = 2x + 4, the original formula = 2x & lt; 4, the original formula = 2x + 4, the original formula = 2x & lt; 4 = 2x & lt; 4, the original formula = 2x & lt; 4, the original formula = 2x & lt; 4, the original formula = 2x & lt; 4, the original formula = 2x & lt; 4 = 2x & lt; 4, the original formula = 2X & lt; 4, the original formula = 2x & lt; 4, the original formula = 2x & lt; 4, the answer is: (1) - 5x-5x-1 (1-5x-5x-1 (1 (1) - 5x-5x-5x-1 (1) (1 (1) - 5x-5x-5x-1 (1 (1) (1 (1) (1) - 5x & lt; (x ≥ 4)

To find the minimum value of | X-1 | + | x + 1 | + | x + 5 |, please use the zero segmentation method

In fact, the zero point method is not used
The meaning of absolute value is the distance from a point on the x-axis to 1, - 1, - 5, so the topic becomes the minimum value of the distance from a point on the x-axis to 1, - 1, - 5, so when x = - 1, the minimum value of | X-1 | + | x + 1 | + | x + 5 | - is 6

How to use the zero segment method in | x + 2 | - | 2x-1 | > 1
I want to know how to find the inequality of absolute value

When there is an absolute value, the zero segment method is to consider it by segment
|X+2|-|2X-1|>1
When x > 0.5,
x＋2－2x＋1>1
3－x>1
－x>-2
x1
x>0
When x1
x－3>1
x>4
When x > 0.5, x0
When X4

2 / 3-x + 1 / 7 = 9 / 14?

2/3-x=9/14-1/7 2/3-x=1/2 x=2/3-1/2 x=1/6

2-1 [3 / 4 (2 / 7 + 5 / 14)]

2-[3/4-(2/7+5/14)]=2-[3/4-(4/14+5/14)]
=2-[3/4-9/14]
=2-[21/28-18/28]
=2-3/28
=1+25/28
=53/28

3\7>( )>2\5; 1\5>( )>1\4

3 / 7 = 15 / 35, 2 / 5 = 14 / 35, obviously there are countless numbers between the two numbers with a difference of 1, 14.1, 14.2, 14.02, 14.1 / 35 = 141 / 350, 14.2 / 35 = 142 / 350 = 71 / 175, 14.2 / 35, and

5 of 7 * 1 of 4 + 3 of 4 * 5 of 7 (2) 1 of 3 + 4 of 7 + 2 of 3 + 1 of 7

&Nbsp; & nbsp; original = 5 / 7 * (1 / 4 + 3 / 4) = 5 / 7 original = 1 + 5 / 7 = 12 / 7

3/1｛7/1[5/1(x+2/3+4)+6]+8｝=3

1/3{1/7[1/5（x+2/3+4）+6]+8}=3
{1/7[1/5（x+2/3+4）+6]+8}=3*3
1/7[1/5（x+2/3+4）+6]+8=9
1/7[1/5（x+2/3+4）+6]=1
1/5(x+2/3+4)+6=7
1/5(x+2/3+4)=1
x+2/3+4=5
x=5-4-2/3
x=1/3
One third means one third

Mathematics problem x + 1 of X + 2 + X + 7 of X + 8 = x + 5 of X + 6 + X + 3 of X + 4
X + 1 of X + 2 + X + 7 of X + 8 = x + 5 of X + 6 + X + 3 of X + 4

X + 1 part X + 2 + X + 7 part X + 8 = x + 5 Part X + 6 + X + 3 Part X + 4 the sum of two X constants on each side of the equation is 10, so we get 1 / x + 7 / x = 5 / x + 3 / X as the identity, so x is any real number not equal to zero