If 2x + 5Y + 4Z = 0, 3x + y-7z = 0, then the value of X + Y-Z is equal to () A. 0b. 1C. 2D

If 2x + 5Y + 4Z = 0, 3x + y-7z = 0, then the value of X + Y-Z is equal to () A. 0b. 1C. 2D

According to the meaning of the question: 2x + 5Y + 4Z = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) 3x + y − 7z = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2), change (2) into: y = 7z-3x, substitute (1) into: x = 3Z, substitute (2) into: y = - 2Z, then x + Y-Z = 3z-2z-z = 0
Given that 2x and 3x-5 are opposite numbers, we can find the value of X
The sum of two opposite numbers is 0
So 2x + 3x-5 = 0
5x=5
X=1
2X = - (3-5) - (3x-5) remove brackets = 5-3x, then add - 3x on both sides of the equal sign, then - x = 5, x = - 5
2x=-(3x-5)=5-3x
X=1
It should be 2x and 3x-5, which are opposite to each other
Then 2x = - (3x-5)
The solution is x = 1
｛6x+5y=25① 3x+4y=20②
②x2-①：3y =15
Y=5
Substituting y = 5 into (1): 6x + 5x5 = 25
6x=0
X=0
6x+5y=25①
3x+4y=20②
1-2 × 2
6x+5y-6x-8y=25-40
-3y=-15
Y=5
It can be obtained by using formula 1;
6x+25=25
X=0
The solution is as follows
x=0 y=5
①×4-②×5
24x-15x=100-100
9x=0
So x = 0
y=(25-6x)/5=5
Solve the equation √ 2x & sup2; + 4 √ 3x = 2 √ 2
√2X2＋4√3X-2√2=0
Using formula method
x=(-4√3±8)/(2√2)
x=-√6±2√2
-√6
√2X²＋4√3X=2√2
Divide both sides by √ 2
X²＋4√3 /√2 X = 2
X²＋2√6 X = 2
X²＋2√6 X + 6 = 2 + 6
(X＋√6)² = 8
(X＋√6) = ±√8
X = ±√8 - √6 = ±2√2 - √6
It can be solved by the root formula
Given / 6x + 5y-25 / + (3x + 4y-20) (3x + 4y-20) = 0, find the value of X and y
Because the sum of these two nonnegative numbers is zero, they can only be zero. So we get the equations 6x + 5y-25 = 0, 3x + 4y-20 = 0. The solution is x = 0, y = 5
Find the equation of the line perpendicular to the line x-3y + 1 = 0 and tangent to the curve y = x ^ 3 + 3x ^ 2-5
Perpendicular to the straight line x-3y + 1 = 0, the slope of the straight line x-3y + 1 = 0 is K1, and the slope of the vertical line L is K2,
Then K1 * K2 = - 1, the vertical equation is 3x + y + M = 0,
If it is tangent to the curve, then calculate the slope of the curve K3 = 3x ^ 2 + 6x, K3 = K2, and get 3x ^ 2 + 6x = - 3, and x = - 1,
If the tangent point is (- 1, - 3), and M = 6 is obtained by taking into the equation L, then the obtained equation is 3x + y + 6 = 0
3x-5y=6 x-4y=-15
Solving equations
3x-5y=6 (1)
x-4y=-15 (2)
From (2), x = 4y-15
Substituting (1)
3(4y-15)-5y=6
12y-45-5y=6
7y=51
y=51/7
x=4y-15=99/7
3x-5y=6.............[1]
x-4y=-15.............[2]
[2]*3:
3x-12y=-45.............[3]
[1]-[3]:
7y=51
y=51/7
x=4y-15=99/7
Equation 2 times 3 minus equation 1
Y = 7 and 2 / 7
Take any equation x and do it yourself
3x-5y=6 （1）
x-4y=-15 （2）
From (2): x = - 15 + 4Y (3)
Substituting (3) into (1) yields:
3 times (- 15 + 4Y) - 5Y = 6
The result is: - 45 + 12y-5y = 6
7y=51
Fifty-one
y= ―
Seven
When x=______ The quadratic function y = x2 + 2x-2 has a minimum value, and its minimum value is______ .
∵ the quadratic function y = x2 + 2x-2 can be reduced to y = (x + 1) 2-3, when x = - 1, the quadratic function y = x2 + 2x-2 has the minimum value of - 3, so the answers are: - 1, - 3
Given that point a (3x-6, 4Y + 15) and point B (5Y, x) are symmetric about X axis, then the value of X + y is ()
A. 0B. 9C. -6D. -12
∵ point a (3x-6, 4Y + 15), point B (5Y, x) are symmetric about the X axis, ∵ 3x-6 = 5Y; 4Y + 15 + x = 0; the solution is: x = - 3, y = - 3, ∵ x + y = - 6, so C
The minimum value of quadratic function y = 2x-x + 1 is
y=2x^2-x+1
=2(x-1/4)^2+7/8
≥7/8
So the minimum value of quadratic function y = 2x ^ 2-x + 1 is 7 / 8
If you don't understand, please hi me, I wish you a happy study!
y=2x²-x+1
=2(x²-1/2x+1/16)-1/8+1
=2(x-1/4)²+7/8
When x = 1 / 4, the minimum value of Y is 7 / 8
Y = 2x-x + 1
=2(x^2-1/2x+1/16-1/16)+1
=2(x-1/4)^2-1/8+1
=2(x-1/4)^2+7/8
When x = 1 / 4, the minimum value of Y is 7 / 8
Y = 2x-x + 1
=2(x^2-1/2x+1/16-1/16)+1
=2(x-1/4)^2-1/8+1
=2(x-1/4)^2+7/8
When x = 1 / 4, the minimum value of Y is 7 / 8
right key
That is to find its vertex formula is 4A / 4ac minus B, so it is 7 / 8
It's easy to learn derivative
Derivation of y = 2x & # 178; - x + 1
When derivative = 0, there is a maximum
y‘=4x-1=0
x=1/4
When x = 1 / 4, there is a maximum and a minimum of 7 / 8