2(x+1)=-3(x-4) solve equations

2(x+1)=-3(x-4) solve equations

2x+2*1=-3x+3*4
2x+3x=3*4-2*1
5x=10
x=10÷5
x=2

3/x+1=4/x+2

3/x+1=4/x+2
3（x+2)=4(x+1)
3x+6=4x+4
4x-3x=6-4
x=2

3 (x + 2) = 4 (x + 1) = how much

2

To solve the fractional equation: (1) 1x − 3 + 2 = x − 43 − X; (2) 2Y2 + 2Y − 3-y − 1Y2 − 9 = − 1y − 1

(1) By multiplying (x-3) on both sides of the equation, we can get 1 + 2 (x-3) = - (x-4) 2, and the solution is x = 3. By testing, x = 3 is an increasing root, so the original equation has no solution. (2) by multiplying (y + 3) (Y-3) (Y-1) on both sides of the equation, we can get 2 (Y-3) - (Y-1) 2 = - (y2-9), and the solution is y = 4. By testing, y = 4 is the solution of the original equation

5x = 3Y, X is 15 less than y, how many are x and y equal
Or you'll be beaten tomorrow

x：y=3：5
therefore
x=15÷(5-3)×3=22.5
y=22.5÷3×5=37.5

Let a be a non-zero vector and λ be a non-zero real number. The correct conclusion in the following is? A, a is opposite to - λ a, B, | - λ a | ≥ | a|
B why not? Even if ramda is - 1, there is a minus sign in front of him. Isn't that positive? Should it be greater than?

λ is a nonzero real number
λ is 0.1
It's not right
B not right

In the RT triangle ABC, AC = BC, there are two e ` f points on the side of AB, making ∠ ECF = 45 degrees, proving AE ^ + BF ^ = EF^

According to the meaning of the title, △ ACB is an isosceles right triangle, CB = Ca, take CB as the edge, make △ CBE '≌ △ CAE outside △ ACB, connect Fe', then CE '= CE, ∠ FCE' = ∠ FCB + ∠ BCE '= ∠ FCB + ∠ ace = 90 ° - ∠ FCE = 90 ° - 45 ° = 45 ° = ∠ FCE, so △ FCE' ≌ △ FCE, Fe '= Fe
In △ FBE ', the ∠ FBE' = ∠ FBC + ∠ CBE '= ∠ FBC + ∠ CAE = 90 ° and be' = AE. The three sides of right angle △ FBE 'conform to be' & sup2; + BF & sup2; = Fe & sup2;, that is AE & sup2; + BF & sup2; = EF & sup2

If the functions f (x) and G (x) are all odd functions defined on R, f (x) = AF (x) + BG (x) + 2 in the interval (0, + ∞), the maximum value is 5,
Finding the minimum value of F (x) in the interval (- ∞, 0)

-1
AF (x) and BG (x) are all odd functions
Let H (x) = AF (x) + BG (x), then H (x) is the maximum value of odd function in the interval (0, + ∞), 3, the minimum value of odd function in the interval (- ∞, 0) - 3, and then add 2 to get - 1

Vector
Given that the midpoint of the sides BC, Ca and ab of the triangle ABC are d (- 3,2), e (5,2) and f (- 1,4), find the coordinates of a, B and C

Let the abscissa of a, B, C be x, y, Z respectively, then x + y = - 1 * 2x + Z = 5 * 2Y + Z = - 3 * 2, solve the system of linear equations of three variables, and get x = 7, y = - 9, z = 3, so the abscissa of a, B, C are 7, - 9,3 respectively. Similarly, the ordinates of a, B, C are 4,4,0 respectively, so the coordinates of a, B, C are (7,4), (- 9

As shown in the figure, the point O is the outer center of △ ABC, and the angle a = 72 ° is used to calculate the degree of the angle BOC

O is the outer center of △ ABC, angle a = 72 °, angle a is the circumference angle, and angle BOC is the center angle, which corresponds to arc BC with angle a, so angle BOC = 2 * angle a = 144 °