The problem of finding the absolute value of mathematics in the first volume of the seventh grade

The problem of finding the absolute value of mathematics in the first volume of the seventh grade

Absolute value is very simple to make clear the negative number, positive number and zero. Zero is neither positive nor negative, and the absolute value of negative number must be equal to positive number, the absolute value of positive number is equal to itself. The absolute value of zero is equal to zero. Pay attention to the sign when calculating the absolute value. Try to open the absolute value with brackets

Seventh grade volume 1 mathematics rational numbers (on absolute values)
The letter X can take innumerable rational numbers, and | x + 2 | has innumerable values. Please tell us whether there are maximum and minimum values in these numbers, whether there are maximum and minimum values in | x + 2 | 7, and the value of X

|X + 2 | has no maximum and a minimum of 0
|Similarly, x + 2 | + 7 has no maximum but a minimum of 7

It is known that a and B are opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to 2. Try to find x ^ 2 + (a + b) - CDX - (- CD) ^ 2010,

It is known that a and B are opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to 2,
Then a + B = 0;
cd=1；
x=±2；
x^2+(a+b)-cdx-(-cd)^2010
=4+0-x-（-1）ˆ2010
=4±2-1
=3±2；
=5 or 1;
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Let a, B and C be positive numbers, and try to prove the inequality: B + Ca + C + AB + A + BC ≥ 6

It is proved that: ∵ a > 0, b > 0, C > 0, ∵ Ba + ab ≥ 2, Ca + AC ≥ 2, CB + BC ≥ 2 ∵ (Ba + AB) + (Ca + AC) + (CB + BC) ≥ 6, that is, B + Ca + C + AB + A + BC ≥ 6

How to prove that the square difference of two sides divided by the value of the other side equals sin (a-b) divided by the square of sinc
Help me

The problem is that the square difference of two sides divided by the square of the value on the other side equals sin (a-b) divided by sinc
Let three sides a, B, C correspond to angles a, B, C and circumcircle radius R
Then a = 2rsina B = 2rsinb C = 2rsinc sin (a + b) = sinc
Then (a ^ 2-B ^ 2) / C ^ 2 = (the square of Sina - the square of SINB) / the square of sinc
=The square of (Sina + SINB) (Sina SINB) / sinc
=The square of 2Sin [(a + b) / 2] cos [(a-b) / 2] * 2cos [(a + b) / 2] sin [(a-b) / 2] / sinc
=The square of 2Sin [(a + b) / 2] cos [(a + b) / 2] * 2cos [(a-b) / 2] sin [(a-b) / 2] / sinc
=The square of sin (a + b) sin (a-b) / sinc
=The square of sincsin (a-b) / sinc
=sin(A-B)/sinC

Given √ 3sin α + cos α = 2 / 3, find cos (2 α - π / 3)

√3sinα+cosα=2sin(a+π/3)=2/3
So sin (a + π / 3) = 1 / 3, cos (a - π / 6) = cos (π / 6-A) = sin [π / 2 - (a + π / 3)] = plus or minus 2 (radical 2) / 3
cos（2a-π/3）=2[cos(a-π/6)]^2-1=7/9

Given cos θ = 13, θ∈ (0, π), then cos (π + 2 θ) equals ()
A. −429B. 429C. −79D. 79

Because sin2 θ + Cos2 θ = 1, cos θ = 13cos (π + 2 θ) = - Cos2 θ = - (Cos2 θ - sin2 θ) = - (2cos2 θ - 1) = 79, D is selected

If cos α = 3sin α, then cos ^ 2 α + 1 / 2sin2 α is equal to

cos^2α+1/2sin2α=cosα/2sinα=1.5

Given that Tana and tanb are the two roots of the equation x ^ 2-4px-3 = 0, find 2cos2acos2b + 2Sin ^ 2 (a-b)
thank you

Tana and tanb are two roots of the equation x ^ 2-4px-3 = 0. Tana + tanb = 4ptana &; tanb = - 3tan (a + b) = (Tana + tanb) / (1-tana &; tanb) = p2cos2acos2b + 2Sin ^ 2 (a-b) = Cos2 (a-b) + Cos2 (a + B) + 1-cos2 (a-b) = 1 + Cos2 (a + b) = 2cos & sup2; (a + b) = 2 /

Mathematical reduction of Cos2 α + 2Sin ^ 2 α

cos2α+2sin^2α=cos²α-sin²α+2sin²α=cos²α+sin²α=1