Calculation (1) (- 1 6 / 7) + (- 1 1 / 7) (2) (- 1 5 / 12) + (+ 1 / 4) (3) (+ 3.2) + (- 4.5) (4) (+ 4 1 / 3) + (5 1 / 2) Quick answer to study If there is no standard answer today The problem of recycling in the past is not accepted

Calculation (1) (- 1 6 / 7) + (- 1 1 / 7) (2) (- 1 5 / 12) + (+ 1 / 4) (3) (+ 3.2) + (- 4.5) (4) (+ 4 1 / 3) + (5 1 / 2) Quick answer to study If there is no standard answer today The problem of recycling in the past is not accepted

I just want the answer

Calculation - 1 + 2 + 3-4-5 + 6 + 7-8-9 + 10 + 11-12. - 2009 + 2010

1

How to solve this equation?

(x+18)÷(48+x)=4/9
9(x+18)=4(48+x)
　　9x+162=192+4x
　　5x=30
　　x=6

Trigonometric function sin7 / 5pai * cos (- 11 / 6pai) + Tan (- 15 / 4pai) * tan7 / 5pai

Simplification of trigonometric function: the original formula = sin (π + 2 π / 5) * cos [- (π + 5 π / 6)] + Tan [- (4 π + π / 4)] * Tan (π + 2 / 5 π). = - sin (2 π / 5) cos (5 π / 6) - Tan (π / 4) * Tan (2 π / 5). = - sin (2 π / 5) * [- cos (π / 6)] - 1 * Tan (2 π / 5). = (√ 3 / 2) sin (2 π

Advanced Algebra: let x1, X2 and X3 be the three roots of the equation x ^ 3-2x ^ 2 + 5x + 2 = 0, and try to calculate the value of (x1) ^ 4 + (x2) ^ 4 + (x3) ^ 4

Notice that X1 + x2 + X3 = 2
x1x2+x2x3+x3x1=5
x1x2x3=-2
(x1) ^ 4 + (x2) ^ 4 + (x3) ^ 4 can be solved by using symmetric polynomial expansion

If a + B = 90 ° then Tana / 2 + tanb / 2 + Tana / 2tanb / 2=

A+B=90°,A/2+B/2=45º
tan(A/2+B/2)=(tanA/2+tanB/2)/(1-tanA/2tanB/2)=1
tanA/2+tanB/2=1-tanA/2tanB/2
tanA/2+tanB/2+tanA/2tanB/2=1-tanA/2tanB/2+tanA/2tanB/2=1

Let P and Q be the points on x-axis and y-axis respectively, and the midpoint m (1, - 2), then the absolute value PQ is equal to?

Method 1
Let P (x, 0) Q (0, y)
∵ midpoint m (1, - 2), ∵,
∴x+0=2,0+y=-4
∴x=2,y=-4
|PQ|=√(x^2+y^2)=√20=2√5
Method 2
|OM|=√(1^2+2^2)=√5
|PQ|=2|OM|=2√5

On the proof of periodic function~
Let f (x) be a periodic function with t (T > 0) as a period, and prove that f (AX) (a > 0) is a periodic function with T / A as a period. Why do we only need to prove that f [a (x + T / a)] = f (AX + T)?

Answer:
The essence of the definition of periodic function is that the value of function is equal when the independent variable is added with a constant,
In this problem, we prove that f (AX) (a > 0) is a periodic function with T / A as period
We only need to prove that the values of X and X + T / a are equal,
Naturally, we only need to prove that f (AX + T) = f [a (x + T / a)]
That is, f [a (x + T / a)] = f (AX + T)

(I) given that the coordinates of the three vertices of △ ABC are a (0,5), B (1, - 2), C (- 6,4), we can find the equation of the line with the height on the edge of BC; (II) let the equation of the line l be & nbsp; (A-1) x + y-2-a = 0 (a ∈ R). If the intercept of the line L on the two coordinate axes is equal, we can find the equation of the line L

(I) the slope of the straight line on the edge of BC is KBC = − 2 − 41 − (− 6) = − 67, the slope of the straight line on the edge of BC is k = 76, the equation of the straight line on the edge of BC is y = 76x + 5, that is, 7x-6y + 30 = 0. (II) let x = 0, y = 2 + A; let y = 0, when a ≠ 1, x = 2 + AA − 1, the intercept of the straight line L on the two coordinate axes is equal, 2 + a = 2 + AA − 1, 2 + a = 0, or A-1 = 1, A-1=- 2, or a = 2, so the linear equation is x + y-4 = 0 or 3x-y = 0

The simple method is 125 times (17 times 18) times 4

=125×4×17×18
=500×18×17
=9000×17
=153000