Given (a + 1) & sup2; + (2b-3) + |c-1 | = 0, find the value / = fractional line of AB / 3C + a-c / b

Given (a + 1) & sup2; + (2b-3) + |c-1 | = 0, find the value / = fractional line of AB / 3C + a-c / b

（a＋1）²＋（2b－3)²＋|c－1|＝0
（a＋1）²=0 ;（2b－3)²=0 ;|c－1|＝0
The solution is a = - 1, B = 3 / 2, C = 1
ab/3c＋a
=(-1)*(3/2)/(3*1-1)
=(-3/2)/2
=-3/4

In △ ABC, the lengths of BC, Ca and ab are a, B and C respectively. (1) prove that a = bcosc + ccosb; (2) if the vector AB * the vector BC + C & sup2; = 0, try
In △ ABC, the lengths of BC, Ca and ab are a, B respectively. (1) prove that a = bcosc + ccosb; (2) if vector AB * vector BC + C & sup2; = 0, try to prove that △ ABC is a right triangle

(1)bcosC+ccosB=BC=a=3acosB
Then CoSb = 1 / 3
(2)accosB=2
ac=6
Cosine theorem
a^2+c^2-2accosB=b^2
a^2+c^2=12
(a-c)^2=0
That is, a = C = 2 times the root 6

The analytic formula of parabola is y = ax & sup2; + BX & sup2; + C, which satisfies the following four conditions: ABC = O, a + B + C = 3, AB + BC + Ca = - 4, a < B < C
(2) whether there is Q on the parabola above the x-axis, so that △ AQC is a right triangle. If there is, ask for the Q coordinate. If not, say the reason

I have the solution to the third question just now
So if you don't understand, the product of the slopes of two straight lines perpendicular to each other is - 1
That is, K1 * K2 = - 1, let Q (T, - T & sup2; + 4)
① When AC ⊥ CQ, (- T & sup2; + 4-4) / T × 2 = - 1, the solution is t = 1 / 2
That is, the coordinates of Q point are (1 / 2,15 / 4)
② When CQ ⊥ AQ, (- T & sup2; + 4-4) / T × (- T & sup2; + 4) / (T + 2) = - 1, the solution is t = 1
That is, the coordinates of Q point are (1,3)

Given X / 2 = Y / 3 = Z / 4, find (XY + 2yz-3xz) / (x ^ 2 + y ^ 2 + Z ^ 2)?

Let X / 2 = Y / 3 = Z / 4 = t, then x = 2T, y = 3T, z = 4T,
So (XY + 2yz-3xz) / (x ^ 2 + y ^ 2 + Z ^ 2)
=(6t^2+24t^2-24t^2)/(4t^2+9t^2+16t^2)
=6/29.

Given X / 2 = Y / 3 = Z / 4, find the value of the algebraic formula x-2y + 3Z / XY + 2yz + 3yZ

There is something wrong with your problem. The general idea is to let x = 2K, y = 3k, z = 4K and substitute it into the original formula

Given that x, y, Z satisfies x + y = 5, Z & # 178; = XY + Y-9, find the value of X + 2Y + 3Z

x+y=5
x=5-y
z^2=xy+y-9
z^2=(5-y)y+y-9
z^2=-y^2+6y-9
z^2=-(y-3)^2
z^2+(y-3)^2=0
So, z = 0, Y-3 = 0
z=0,y=3
x=5-y=5-3=2
x+2y+3z=2+2*3+3*0=8

Given X / 2 = Y / 3 = Z / 4, find X & # 178; + 2Y & # 178; + 3Z & # 178; / XY + YZ ZX
By the way, this chapter is about fractions. Please use "fractions" to solve it,

Let X / 2 = Y / 3 = Z / 4 = k, then
x=2k,y=3k,z=4k
So x & # 178; + 2Y & # 178; + 3Z & # 178; / XY + YZ ZX
=4k²+18k²+48k²/6k²+12k²-8k²
=70k²/10k²
=7

Let z = f (x, y) satisfy X & # 178; + 2Y & # 178; + 3Z & # 178; + xy-z-9 = 0, the second partial derivative of Z to X is obtained

2x+6z ∂z/∂x +y-∂z/∂x=0∴(6z-1)∂z/∂x=-2x-y ∂z/∂x=(2x+y)/(1-6z)∂²z/∂x²= [2(1-6z) -(2x+y) (-6∂z/∂x)]/(1-6z)² ...

(3) factorization: (1) 4A ^ 2 + A + 1 / 16 (2) 9 (M + n) ^ 2 + 6m + 6N + 1 (3) (a + b) ^ 2-4 (a + B-1)

1) 4A ^ 2 + A + 1 / 16 = (2a + 1 / 4) &;
（2）9（m+n)^2+6m+6n+1=9（m+n)²+6(m+n)+1=(3m+3n+1)²
（3）（a+b)^2-4(a+b-1)=(a+b)²-4(a+b)+4=(a+b-2)²

It is known that the quadratic trinomial 2x ^ 2-3x + m cannot decompose the factor in the range of real number, so the value range of M can be obtained

According to the zero point expression of quadratic function f (x) = a (x-x1) (x-x2), when the quadratic trinomial 2x ^ 2-3x + m cannot decompose the factor in the range of real number, then the equation 2x ^ 2-3x + M = 0 has no real root
So the discriminant is 3 ^ 2-4 * 2m9 / 8