Are 10 + x > 5 and 0.4x = 8 equations

Are 10 + x > 5 and 0.4x = 8 equations

They are all equations. The former belongs to inequality equation, and the latter belongs to equality equation~

When Xiao Ming solved the equation T ^ 2 + 2T = 8 by factorization
So the original equation is changed to t (T + 2) = 2 * 4, so t = 2, t + 2 = 4, so the root of the original equation is T1 = 2,
T2 = 2. When Xiao Hong solves the equation 2x (x + 1) = x + 1, we get 2x = 1 by deleting x + 1 from both sides. So x = 1 / 2. Is the solution of Xiao Ming and Xiao Hong correct? If there is any error, please give the correct solution

The answers given by Xiao Ming and Xiao Hong are incorrect. The correct answers are as follows:
t^2+2t=8
t^2+2t-8=0
(t-4)(t+2)=0
t=4 or t=-2
2x(x+1)=x+1
2x(x+1)-(x+1)=0
(x+1)(2x-1)=0
x=-1 or x=-1/2

Knot equation x (2x-5) = 4x-10

x(2x-5)=4x-10
Let's start
2x^2-5x=4x-10
Moving items
2x^2-9x+10=0
Ten character multiplication
2 -5
1 -2
（2x-5)(x-2)=0
x1=2.5,x2=2
So the solution of the equation is X1 = 2.5, X2 = 2

There are several ways to solve equation by factorization
Quick return

This method is called "matching method". 2. Factorization cross multiplication cross multiplication is a basic method of factorization

The following equation is solved by factorization: (2x + 3) (2) = 24x

4x^2+12x+9-24x=0
4x^2-12x+9=0
(2x-3)^2=0
x=1.5

2T & # 179; - 6T & # 178; + 6t-1 = 0 for t

Wrong question, no answer! T (T & # 178; - 6T + 6) + (t-1) (T & # 178; + T + 1) = 0, no solution

1/2+2/3+...+55/56

Let me give you an example. I hope you can understand 1/2+2/3+3/4+.+99/100= (1-1/2)+(1-1/3)+(1-1/4)+.+(1-1/100） =55-(1/2+1/3+1/4+...+1/100） 1/2+1/3+…… +1 / 100 is a harmonic series term, there is no formula, only approximate value, if you want to use Excel function

A particle moves in a straight line along the x-axis, and its motion equation is x = 2 + 6T & # 178; - 2T & # 179;,
Calculate: ① the displacement of the particle within 4S after the start of the motion; ② the distance the particle passes in this time; ③ the velocity and acceleration of the particle when t = 4S

Analysis: 1. When T1 = 0, the position is X1 = 2 meters; when T2 = 4 seconds, the position is x2 = 2 + 6 * 4 ^ 2-2 * 4 ^ 3 = - 30 meters, so the displacement of the particle in 4 seconds is s = x2-x1 = (- 30) - 2 = - 32 meters

120 / x-120 / (1.5x) = 1, solve the equation and test it

120*1.5/(1.5x)-120/(1.5x)=1
(180-120)/(1.5x)=1
1.5x=60
x=40
120/40-120/(1.5*40)=3-2=1

The value of 2T & # 179; - 6T & # 178; + 6t-1 = 0 t

We use the difference cube formula: (a-b) &# 179; = A & # 179; - 3A & # 178; B + 3AB & # 178; - B & # 179;
2t³-6t²+6t-1=0
2(t³-3t²+3t-1)+1=0
2(t-1)³+1=0
(t-1)³=-1/2
T = (- 1 / 2) open cubic + 1