Let the real number x satisfy the square of equation (X-2) + (KX + 2) = 4, and find the maximum value of K

Let the real number x satisfy the square of equation (X-2) + (KX + 2) = 4, and find the maximum value of K

Square of (X-2) + (KX + 2) = 4
The results show that: (1 + K & sup2;) x + (4k-4) x + 4 = 0
Because x is a real number
△=-32k≥0
Then K ≤ 0
The maximum value of K is 0

On the equation (a-6) of X, where the square of x-8x + 6 = 0 has real roots, then the maximum value of integer a is
A.6
B.7
C.8
D.9

△=64-24(a-6)≥0
24（a-6)≤64
3(a-6)≤8
3a≤26
a≤26/3
So the maximum value of integer a is 8

3 (x squared minus 2x) squared minus 21 (x squared minus 2x) minus 24 factorization requires a process

3（x^2-2x）^2-21(x^2-2x)-24
=3 {(x^2-2x)^2-7(x^2-2x)-8 }
=3(x^2-2x-8)(x^2-2x+1)
=3（x-4）(x+2)(x-1)^2
^2 is the square

A large cube is made up of 27 small cubes whose edges are 1 cm long. The surface area of the cube is______ And the volume is______ ．

The edge length of the assembled cube is 3cm. Surface area: 3 × 3 × 6 = 36 (square centimeter) volume: 3 × 3 × 3 = 27 (cubic centimeter) answer: the surface area of the cube is 36cm square and the volume is 27cm cubic. So the answer is: 36cm square, 27cm cubic

How do 7 and 10 and 9 and 6 equal 24

This problem has no result, not all combinations can work out 24

It is known that all items of the sequence {an} are positive numbers, the first n terms and Sn satisfy Sn = 1 / 6 (an + 1) (an + 2), and A2, A4, A9 are equal proportion sequence
(1) (2) BN = an * an + 1, TN is the sum of the first n terms of the sequence {BN}, T2N
All lowercase n are subscripts, and N in S = 1 / 6 (an + 1) (an + 2) in the known condition is subscript + 1 and + 2, which are not subscripts but additive numbers
N and N + 1 in BN = an * an + 1 in question 2 are subscripts

All of a sudden, what we have just done is gone. Let's start with Sn = 1 / 6 (an + 1) (an + 2), a1 + A2 + a3 +... + an = 1 / 6 (an + 1) (an + 2), 6 (a1 + A2 + a3 +... + an) = (an + 1) (an + 2), 6 (a1 + A2 + a3 +... + an-1) = A & sup2; n-3an + 2 = (an-1) (An-2), 6 (a1 + A2 + a3 +... + an-1) = 6sn-1, 6S (n-1) = (a (n-1)

The perimeter of a square with a side length of a decimeter is () and the area is（

The perimeter of a square with a side length of a decimeter is (4a) decimeter, and the area is (A & # 178;) square decimeter

Simple calculation: (27-23) × 12

(27-23)×12
=[（25+2﹚﹣﹙25－2）]×2×6
=[25＋2－25＋2]×2×6
=[2＋2]×2×6
=4×2×6
=8×6
=48

Let f (x) be an increasing function of R hill, f (x 1) + F (x 2) > F (x 3) + F (x 4), then x 1 + x 2 > x 3 + x 4
Why, I want a clear and easy answer!

For example: if f (x) = x ^ 3, X1 = 0, X2 = 4, X3 = 2, X4 = 3, then f (x1) + F (x2) = 0 + 4 ^ 3 = 64, f (x3) + F (x4) = 2 ^ 3 + 3 ^ 3 = 8 + 27 = 35, f (x1) + F (x2) > F (x3) + F (x4), but X1 + x2 = 4

Using 1.2 square meters of tiles to lay the living room needs 108 pieces. If you use 1.8 square meters of tiles to lay the living room, how many pieces do you need?

1.2×108÷1.8
=129.6÷1.8
=72 (pieces)
First calculate the living room area, and then calculate the number of blocks