To solve the equation, 11 / 24-x = 1 / 5, 6x + 1.5x = 15, 3x-4 / 3x = 1.5, 4x-1.5 * 2 / 3 = 5 / 7

To solve the equation, 11 / 24-x = 1 / 5, 6x + 1.5x = 15, 3x-4 / 3x = 1.5, 4x-1.5 * 2 / 3 = 5 / 7

The first question is 11 / 24-x = 1 / 5;
x=55/120-24/125;
x=21/125;
Question 2 6x + 1.5x = 15
7.5x=15;
x=2;
Question 3: 3x = 1.5
0.25x=1.5;
x=6;
Question 4 4x-1.5 * 2 / 3 = 5 / 7;
4x=1+5/7;
4x=12/7;
x=3/7;
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It is known that the equation a (3x + 2) + B (- 2x + 3) = 5x + 12 has infinite solutions, then a = B=
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4 (3.1-x) = 2.2x, 12-2x = 1.5x + 5, X △ 2 + 6.25 = 3x

4 (3.1-x) = 2.2x, 12.4-4x = 2.2x, 6.2x = 12.4x = 2

The difference between the polynomial 8xy-x ^ 2 + y ^ 2 and the polynomial m is - 2x ^ 2 + 2Y ^ 2, M =?

The difference between the polynomial 8xy-x ^ 2 + y ^ 2 and the polynomial m is - 2x ^ 2 + 2Y ^ 2,
M=8xy-x^2+y^2 -(-2x^2+2y^2)
=8xy-x^2+y^2+2x^2-2y^2
=x^2+8xy-y^2

What are the funny English words?

funny

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, then A6 + A7 > 0 is the ()
A. Sufficient but not necessary condition B. necessary but not sufficient condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

Let P: A6 + A7 > 0, Q: S9 ≥ S3 simplify, P: 2A1 + 11d > 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Q: s9-s3 ≥ 0, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 9a1 + 36d - (3A1 + 3D) ≥ 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

F (x) is defined on R, f (x + y) = f (x) + F (y) + 2XY. It is proved that if f '(0) exists, then the function can be derived at any point, and f' (x) can be obtained

It is proved that: let x = y = 0, then: F (0 + 0) = f (0) + F (0) + 2 * 0 * 0, that is, f (0) = 0 for any point x on R, let t be a tiny offset on it, T - > 0, f (x + T) - f (x) = f (x) + F (T) + 2 * x * T-F (x) = f (T) + 2 * t * x, then the derivative of F (x) on X is: [f (x + T) - f (x)] / T = [f (x) + 2 * t * x] /

2000 by 1 / 2001, 2001 by 1 / 2002, 2007 by 1 / 2008,

2000 / 2008
From 2001 to 2007, it can be reduced from denominator numerator

What is the solution set of 3x's square-x-2 = 0,

3x^2-x-2=0
(3x+2)(x-1)=0
x=-2/3 or 1

Finding the extremum of function f (x, y) = XY (2-x-y)

If f'x = y (2-x-y) - xy = y (2-2x-y) = 0, then y = 0 or y = 2-2xf'y = x (2-x-y) - xy = x (2-x-2y) = 0, then x = 0 or x = 2-2y, then the solution has the following groups (0,0), (2,0), (0,2), (2 / 3,2 / 3) a = f "XX = - 2yb = f" xy = 2-2x-2yc = f "YY = - 2x in (0,0), AC-B ^ 2 = 0-2 ^ 2 = - 4