If k x is not a product of K x and K X-1?

If k x is not a product of K x and K X-1?

(x-1)(2-kx)
=2x-kx*x-2+kx
=(2+k)x-k*x^2-2
Because it doesn't contain the first term of X
SO 2 + k = 0
k=-2

It is known that the product of the quadratic trinomial 2x ^ 2-3x + 1 and ax ^ 2 + BX + 1 does not contain x ^ 3 term, and does not contain x term to find the value of coefficient a, B

(2x^2-3x+1)(ax^2+bx+1)
=2ax^4+2bx^3+2x^2-3ax^3-3bx^2-3x+ax^2+bx+1
=2ax^4+()x^3+(2-3b+a)x^2-(3-b)x+1
Because the product doesn't contain x ^ 3 terms, and it doesn't contain x terms
So 2b-3a = 0, 3-B = 0
So B = 3, a = 2

It is known that the equation 2x2 + kx-1 = 0 (1) Proof: the equation has two unequal real roots; (2) If one root of the equation is - 1, find the other root and K value

It is proved that: (1) ∵ a = 2, B = k, C = - 1
∴△=k2-4×2×（-1）=k2+8，
∵ no matter what value k is taken, K2 ≥ 0,
ν K2 + 8 > 0, that is △ > 0,
The equation 2x2 + kx-1 = 0 has two unequal real roots
(2) Substituting x = - 1 into the original equation, 2-k-1 = 0
∴k=1
The original equation is changed into 2x2 + X-1 = 0,
The solution is: X1 = - 1, X2 = 1
The other root is 1
2．

If | x + 2 | + √ (radical x + Y-1) is equal to 0, find the value of the square of the algebraic expression (| 2) x + (2 / 3) y-2x-1.5x + (the square of | 3)

If one is greater than 0, the other is less than 0
So both are equal to zero
So x + 2 = 0, x + Y-1 = 0
x=-2
y=1-x=3
(2) x + (2 / 3) the square of y-2x-1.5x + (the square of
Just to add, ask me

calculation (4x^2-2x^3+6x)/(-2x)-(x-1)^2 y^2-4/y^2-y-6)+(y+2/y-3)/(y+1/y-3)

The original formula = - 2x + x ^ 2-3-x ^ 2 + 2x-1
=-4
The original formula = (y + 2) (Y-2) / (Y-3) (y + 2) + (y + 2) / (y + 1)
=(y-2)/(y-3)+(y+2)/(y+1)
=(y^2-y-2+y^2-y-6)/(y-3)(y+1)
=(2y^2-2y-8)/(y-3)(y+1)

3x-20 + 6x-2 = 8x-10 + 2x, The sum of three consecutive odd numbers is 15 more than the sum of two even numbers________

3X-20+6X-2=8X-10+2X
9X-22=10X-10
X=-12
Let the first odd number be X
Then x + X + 2 + X + 4 = x + 1 + X + 3 + 15
3x + 6 = 2x + 19
x=13
So the five numbers are 13, 14, 15, 16, 17

To solve the problem by equation must be the first mathematical application of a single variable equation Fill a measuring cylinder with 3 cm in diameter and 22 cm in height with water, and then pour the water into a beaker with a diameter of 7 cm and a height of 9 cm. Can it be completely filled? If not, how high is the water left in the cylinder? If it can, calculate the height of the water surface in the cup If the beaker is filled with water and poured into the measuring cylinder, can it be filled? If not, how high is the water left in the beaker

(1) Let the height in the cup be xcm. Because the water is invariable, that is, the volume of water in the two containers remains unchanged
1/4×3^2×22=1/4×7^2×x
The solution is x = 198 / 49
Because 198 / 49 is less than 9, it can be installed. The height in the cup is 198 / 49
(2) Same as above,
Let the height in the cup be xcm. Because the water is invariable, that is, the volume of water in the two containers remains unchanged
1/4×3^2×x=1/4×7^2×9
X = 49
Because 49 ＞ 22, it can not be filled. If the water in the cup is still xcm high
Then 1 / 4 × 3 ^ 2 × 22 = 1 / 4 × 7 ^ 2 × (9-x), x = 243 / 49

The main process sin (α + β) / cos α cos β = Tan α + Tan β sin(α+β)/cosαcosβ=tanα+tanβ sin(α+β)cos(α-β)=sinαcosα+cosβcosβ

Just use the sum of two corners
sin(a+b)/cosacosb
=(sina*cosb+cosa*sinb)/cosa*cosb
=sina/cosa+sinb/cosb
=tana+tanb
Use formula expansion, direct multiplication is OK
sin(α+β)cos(α-β)
=(sinacosb+cosasinb)(cosacosb+sinasinb)
=sinacosa*(cosb)^2+(sina)^2*sinbcosb+(cosa)^2*sinbcosb+
(sinb)^2*sinacosa
=sinacosa[(sinb)^2+(cosb)^2]+sinbcosb[(cosa)^2+(sina)^2]
=sinacosa+sinbcosb

Given the function f (x) = (2 x power - 1 / 1 + 1 / 2), find the definition domain of function f (x) It is proved that f (x) is less than 0

F (x) = (x power of 2 - 1 / 1 + 1 / 2)
That is, f (x) = 1 / (2x-1) + 1 / 2
Find the definition domain of function f (x)
If 2x-1 ≠ 0, 2x ≠ 1, then x ≠ 0
Supplement: it is proved that f (x) is less than 0
f(x)=1/(2x-1)+1/2
f(-x)=1/(2-x-1)+1/2=1/(1/2x-1)+1/2=-f(x)
So f (x) is an odd function

Given x (x-1) - (x2-y) = - 2, find x2 + Y2 2 − XY

∵x（x-1）-（x2-y）=-2，
∴x2-x-x2+y=-2，
∴x-y=2，
∴x2+y2
2-xy=x2+y2−2xy
2=(x−y)2
2=2．