X-8 / 5x = 48 solution equation

5x = 48
3x of 8 = 48
X = 48 △ 3 / 8
X = 8 / 48x3
x= 128

Let the line 2x-y + 1 = 0 intersect the ellipse x2 / 3 + Y2 / 4 = 1 at two points a and B. find the coordinates of the midpoint m of the line AB and the length of the line ab?

When the line 2x-y + 1 = 0, we get y = 2x + 1, and substitute it into the ellipse x2 / 3 + Y2 / 4 = 1, we get 16x2 + 12x-9 = 0,
Let a (x1, Y1) B (X2, Y2), then X1 + x2 = - 12 / 16 = - 3 / 4, x1 · x2 = - 9 / 16, Y1 + y2 = (2x1 + 1) + (2x2 + 1) = 1 / 2
So m (- 3 / 8,1 / 4), ab = √ (1 + K2) ×| x1-x2 | = √ 5 × √ [(x1 + x2) 2-4x1x2] = 15 / 4

For decimal complement: for example -0.0110, the number of digits is eight

The complement of 8-digit fixed-point number is used to represent pure decimal, the highest bit represents sign bit, and the lower 7-digit represents absolute value
The highest digit is 1, which means a negative decimal. The original code of the absolute value of the lower 7 digits is 011000. If you add 1 to the inverse, you get 1010000. If you add sign digit 1, you get 11010000. That is to say, the 8-digit binary complement of -0.011b is 11010000

Given that at least one of the three real numbers a, B and C is not equal to 1, try to compare the sizes of a & sup2; + B & sup2; + C & sup2; and 2A + 2B + 2c-3
Such as the title, know the process of speaking,

a²+b²+c²-(2a+2b+2c-3)=(a-1)^2+(b-1)^2+(c-1)^2＞0
So a & sup2; + B & sup2; + C & sup2; > 2A + 2B + 2c-3

The function y = f (x) is differentiable in the interval (0, + ∞), and the derivative is a decreasing function. Let y = f (x) be the tangent equation of the curve at the point (x0, f (x0)), and let g (x) = KX + M
(I) m is represented by x0, f (x0) and f '(x0);
(II) proof: when;
(III) if the inequality about X is constant, where a and B are real numbers,
Find the value range of B and the relationship between a and B

2. We can't see formula III
solution
(1) G (x) is a tangent equation, so it can be expressed as
g(x) = f'(x0) (x-x0) + f(x0)
Compared with G (x) = KX + M
m = f(x0) - x0 f'(x0)

It is proved that: (y + z-2x) 3 + (Z + x-2y) 3 + (x + y-2z) 3 = 3 (y + z-2x) (Z + x-2y) (x + y-2z)

In this paper, we prove that: let y + Z + z-2x = a, let y + Z + z-2x = a, 1, Z + x-x-2y = B, 2, x + y-2z = C, 3, the equation to be proved will be changed to A3 + B3 + C3 = 3ABC. We associate the multiplication formula: A3 + B3 + B3 + c3-3-3abc = (a + B + B + C) (A2 + B + B2 + C2 + c2-c2-ab-ab-bc-bc-ca) (a + B + B + B + B + B (A2 + B2 + B2 + B2 + c2-c2-ab-bc-bc-ca) (A2 + B2 + B2 + B2 + C2 + c2-c2-c2-ab-ab-bc-bc-bcbcbcca) (a + A2 + B + B + B + B + B2 + B2 + B2 + C2 + C2 + C2 + C2 + C2 + C2 + C2 + c2-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-the results show that: 1

Using other methods to prove congruent right triangle, which triangle does not need to be written as RT triangle?

All right
Write it or not

5% X - 2. 5% x = 9

12.5%X－2.5%X=9
0.1X=9
X=9/0.1
X=90

Let vector groups A1, A2, A3 be linearly independent. It is proved that vector groups a1 + a3, a1-2a3, A2 + a3 are also linearly independent

k1(a1+a3)+k2(a1-2a3)+k3(a2+a3)=0
=> (k1+k2)a1+k3a2+(k1-2k2+k3)a3=0
=> k1+k2=0 (1) and
k3=0 (2) and
k1-2k2+k3=0 (3)
from (3) and (2)
k1-2k2 = 0 (4)
(1)-(4)
3k2=0
=> k2 =0
from (1)
=> k3=0
=>A1 + a3, a1-2a3, A2 + a3 are linearly independent

C + + uses dichotomy to find the approximate root of equation x3-x-1 = 0 in [1.0,1.5]
The error should be less than 1e-5
Tips:
(1) First, we take two rough solutions of the equation f (x), X1 = 1.0 and X2 = 1.5;
(2) If the sign of F (x1) is opposite to that of F (x2), then the equation f (x) = 0 has at least one root in the interval [x1, X2];
(3) Let X3 = (x1 + x2) / 2, if f (x3) = 0, then X3 is the solution of the equation; otherwise, in X1 and X2, if the same sign as f (x3) is omitted, the root is in the interval composed of X3 and the other rough solution without rounding;
(4) Repeat (3) until the difference between xn and XN-1 meets the required error, xn is the approximate root of equation f (x)

#include
using namespace std;
double f(double x)
{
\x09return x*x*x - x - 1;
}
int main()
{
\x09double left = 1;
\x09double right = 1.5;
\x09double mid;
\x09while(right-left > 1e-5)
\x09{
\x09\x09mid = (left + right) / 2;
\x09\x09if(f(mid)==0)
\x09\x09\x09break;
\x09\x09if(f(mid)*f(left)>0)
\x09\x09\x09left = mid;
\x09\x09else
\x09\x09\x09right = mid;
\x09}
\x09cout

X-8 / 5x = 48 solution equation