Comma expression (a = 3 * 5, a * 4), the value of a + 15 is?

Comma expression (a = 3 * 5, a * 4), the value of a + 15 is?

Comma expression, its solution process is expression 1 first, then expression 2, the whole expression value is the value of expression 2, for example: (3 + 5,6 + 8) value is 14
A = 3 * 5; 15 is assigned to a variable;
Equivalent to (15 * 4,15 + 15) according to the principle of comma expression: so the value of the question should be 30

C + + expert please enter comma expression (x = 4 * 5, X * 5), the value of X + 25 is () (a) 25 (b) 20 (c) 100 (d) 45
Why do I think it's 125!

Do you think x should be equal to 100 after X * 5?
Look at the program first
Your expressions are not assigned values. Let me give you an expression with the same format
D = ((a, b), c) you don't give d here, a is your x = 4 * 5, B is your X * 5, C is your X + 25
Remember, comma operation is an expression operation from left to right, which is just the result of the whole comma expression and the last expression result, namely d = C
According to the operation rules, the operations are as follows:
The expression of a is x = 4 * 5. After running, x = 20
B expression, that is, after X * 5 runs: there is no meaning. Here, the value of X will not change at all, but the result of (a, b) will be equal to 100, but our final result is determined by C, so the execution of B does not change the variables in the program
C expression is x + 25 after running: then the final result is d = ((20100), 20 + 25)
Obviously the end result is 45

Calculate the square of (3m-2n Square)

Square of (3m-2n Square)
=The fourth power of 9m & # 178; - 12mn & # 178; + 4N

It is known that three non-zero real numbers a, B, C form an arithmetic sequence, and a ≠ C. It is proved that 1 / A, 1 / B, 1 / C cannot be an arithmetic sequence

Three nonzero real numbers a, B and C are known to form an arithmetic sequence. We have a + C = 2B and divide 1 / A, 1 / B and 1 / C into 1 / A + 1 / C = (BC + AB) / ABC = (a + C) * B / ABC = 2B * B / ABC1 / B + 1 / b = 2Ac / ABC. If 1 / A, 1 / B and 1 / C are arithmetic sequences, then 1 / B + 1 / b = 1 / A + 1 / C needs to satisfy 2Ac / ABC = 2B * B / abcac = b * B

For a triangle test field, it is necessary to divide the test field into four small plots with equal area, plant four different fine varieties, design more than three different division schemes, and give explanations

As shown in the figure:

Given the points m (1,3), n (5, - 2), if there is a point P on the x-axis so that | pm-pn | is the maximum, then the coordinate of point P is______ ．

Let m (1,3) be a symmetric point m '(1, - 3) about X axis, and let m' n intersect X axis at point P, then point P is the solution. Let the analytic expression of M 'n be y = KX + B, and substitute M' (1, - 3), n (5, - 2) into − 3 = K + B − 2 = 5K + B, then k = 14, B = - 134, so the analytic expression of this function is y = 14x-134, when y = 0, x = 13, so the coordinates of point P are (13, 0)

Given a ^ 2 + A-1 = 0, find the value of (a ^ 2 + 3) (a + 2) - 3-3a

a^2+a-1=0
a^2+a=1
(a^2+3)(a+2)-3-3a
=a^3+2a^2+3a+6-3-3a
=a^3+2a^2+3
=a^3+a^2+a^2+3
=a(a^2+a)+a^2+3
=a+a^2+3
=1+3=4

It is known that in rectangular trapezoid ABCD, ad ‖ BC, AB is the diameter of a circle, and ab = AD + BC, it is proved that CD is the tangent of a circle
② The tangents of the circles passing through the two ends of the line are parallel to each other

Let the midpoint of AB be o, that is, the center of the circle, and the midpoint of CD be p, connecting Op,
Then OP is the median line of ladder ABCD
So OP = 1 / 2 (AD + BC) = 1 / 2Ab = OA, and op / / ad
Because of ad ⊥ CD, Op ⊥ CD, OP is the distance from the center of the circle to CD
Then the distance from the center of the circle to CD is equal to the radius, CD is the tangent of the circle

If the real roots of the equations x ^ 5 + X + 1 = 0 and X + x ^ (1 / 5) + 1 = 0 are a and B respectively, then a + B=__

Let x ^ 5 + X + 1 = 0 be -- (1)
Note x + x ^ (1 / 5) + 1 = 0 - (2)
Let T1 be the root of (1), then T1 ^ 5 + T1 + 1 = 0
The deformation is (T1 ^ 5) + (T1 ^ 5) ^ (1 / 5) + 1 = 0
This equation shows that T1 ^ 5 is the root of equation (2)
Note T2 = T1 ^ 5
Obviously, T1 + T2 = T1 + T1 ^ 5 = - 1
In conclusion, equation (1) has n roots, equation (2) must have n roots, and the sum of these 2n roots is - n
By derivation, we can judge that equation (1) has only one real root
The sum of the real roots of the equations x ^ 5 + X + 1 = 0 and X + x ^ (1 / 5) + 1 = 0 is - 1

Quadrilateral ABCD inscribed in circle O, diagonal AC, BD intersected in E, AE = CE, ab = √ 2ae, BD = 2 times root 3, calculate the area of quadrilateral ABCD

Let AE = x, then CE = x, ab = √ 2x, AC = 2x, because BD = 2 √ 3, BF = 4, so f = 60 °, then BCD = 60 ° because AB: AE = √ 2, AC: ab = √ 2, so AB: AE = AC: ab, so △ Abe ∽ ACB, so ∠ Abe = ∠ ACB, because ∠ Abe = ∠ ACD, so ∠ ACB = ∠ ACD = 3