0.3 x-1-0.5 x + 2 = - 4

0.3 x-1-0.5 x + 2 = - 4

10(x-1)/3-2(x+2)=-4
10(x-1)-6(x+2)=-12
10x-10-6x-12=-12
4x=10
x=5/2

(x + 4) / 0.2 - (x-1) / 0.5 = 1?

Go to the denominator first
5（X+4）-2（X-1）=1
5X-2X+20+2=1
3X=-21
X=-7

4^(x+1)-2(6)^(x)-6(9)^(x)=0
Find x

4 ^ (x + 1) - 2 (6) ^ (x) - 6 (9) ^ (x) = 0 is transformed into: 4 * 2 ^ (2x) - 2 * (2 ^ x) * (3 ^ x) - 6 * 3 ^ (2x) = 0, let a = 2 ^ x, B = 3 ^ x, the original formula becomes: 4 * a ^ 2-2 * a * B-6 * B ^ 2 = 02 * a ^ 2-A * B-3 * B ^ 2 = 0 (a + b) * (2a-3b) = 01) a + B = 0, then 2 ^ x + 3 ^ x = 0 has no solution 2) 2a-3b = 0, then 2 * 2 ^ x-3 * 3 ^ x =

If the equation x ^ 2-2x + A-8 = 0 has two real roots x1, X2, and x1 ≥ 3, X2 ≤ 1, find the range of A

Two unequal real roots have (- 2) ^ 2-4 (A-8) = 36-4a
a＜9
X1 is greater than or equal to 3, X2 is less than or equal to 1, and the opening of the function is upward
f（1）＝1-2+a-8＝a-9＜0
f（3）＝9-2*3+a-8＝a-5＜0
So a < 5

Find the maximum area of the inscribed rectangle of the ellipse x ^ 2 + 4Y ^ 2 = 8 and the coordinates of the four vertices of the rectangle

By using affine transformation x = x, y = 1 / 2Y, the inscribed square of circle x ^ 2 + y ^ 2 = 8 with its vertex (± 2, ± 2) is transformed into a rectangle with its vertex (± 2, ± 1) and area 4 * 2 = 8

300 exercises for solving equations

7(2x-1)-3(4x-1)=4(3x+2)-1; (5y+1)+ (1-y)= (9y+1)+ (1-3y); [ ( )-4 ]=x+2; 20%+(1-20%)(320-x)=320×40% 2(x-2)+2=x+1 2(x-2)-3(4x-1)=9(1-x) 11x+64-2x=100-9x 15-(8-5x)=7x+(4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2...

If the product of polynomial MX ^ 2 + nx-1 and X-2 is a cubic binomial of X, find M,

(mx^2+nx-1)(x-2)
=mx^3+(n-2m)x^2-(2n+1)x+2
Because the original formula is cubic binomial
So n-2m = 0 and 2n + 1 = 0
The solution is n = - 1 / 2, M = - 1 / 4

If f (x) = x − 1 x, then the zero of G (x) = f (4x) - x is zero______ ．

∵ f (x) = x − 1X, ∵ f (4x) = 4x − 14x, let g (x) = f (4x) - x = 0, that is 4x − 14x − x = 0, the solution is x = 12, so the answer is 12

Mathematical calculation and solving equation
Calculation: (1) (2 √ 3-3 √ 2) ^ 2
Solution equation: (1) 2x (x-1) = 3 (x-1)
(2)1/2x^2-4√2x-1=0

（1）（2√3-3√2）^2=12-12√6+18=30-12√6
Solution equation: (1) 2x (x-1) = 3 (x-1)
(2x-3)(x-1)=0
x1=3/2,x2=1
(2)1/2x^2-4√2x-1=0
The equation is not clear

The tangent equation of curve y = x ^ 2-5x + 1 at point (2, - 1) is

y=x²-5x+1
The derivation is y '= 2x-5
So when x = 2, y '= 2 × 2-5 = - 1
That is, the tangent slope at point (2, - 1) is - 1
From this we can write out the point oblique equation of the straight line
y-(-1)=-1×(x-2)
It is reduced to: x + Y-1 = 0