3 (5x + 20) = 45x to solve the equation

3 (5x + 20) = 45x to solve the equation

3(5x+20)=45x
===> 5x+20=45x÷3
===> 5x+20=15x
===> 15x-5x=20
===> 10x=20
===> x=2

On the univariate quadratic equation x & # 178; - TX + 2T = 0 of X, if one root is 4, find the value of the other root and t

Take x = 4 into the original formula. 16-4t + 2T = 0, t = 8
Then the original formula = x & # 178; - 8x + 16 = (x-4) &# 178; = 0
There are two identical real roots x = 4

Solve the equation x + 3 / 56 + 29-x / 8 = 1

x+3/56+29-x/8=1
x -x/8 = 1 - 29 - 3/56
7 / 8 x = - 28 and 3 / 56
X = - 32 and 3 / 49

Given one root 2 - √ 2 of the quadratic equation (t-1) x & # 178; - 2tx + 2t-2 = 0 about X, find the other root and the value of T

x1x2=(2t-2)/(t-1)=2
x1=2-√2
So x2 = 2 / (2 - √ 2)
That is, X2 = 2 + √ 2
x1+x2=4=2t/(t-1)
2t=4t-4
t=2

1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90=?
1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
=1/2-1/10
=2/5

1 / 6 + 1 / 12 + 1 / 20 + 1 / 30 + 1 / 42 + 1 / 56 + 1 / 72 + 1 / 90 = 1 / 2 * 3 + 1 / 3 * 4 + 1 / 4 * 5 + 1 / 5 * 6 + 1 / 6 * 7 + 1 / 7 * 8 + 1 / 8 * 9 + 1 / 9 * 10 = 1 / 2-1 / 3 + 1 / 3-1 / 4 + 1 / 4-1 / 5 + 1 / 5-1 / 6 + 1 / 6-1 / 7-1 / 7 + 8-1 / 9-1 / 10 = 1 / 2-1 / 10 = 2 / 5. The solution is correct

It is known that K is an integer, and the cubic power of X + the square of 3 * x - 3x + K has a factor X + 1 Urgent,
Given that K is an integer and that the cubic power + 3 * (the square of x) - 3x + k of X has a factor X + 1, then k =?
The other is the quadratic factor, which is?

Dizziness
set up
Another one (AX ^ 2 + BX + C)
Then (AX ^ 2 + BX + C) (x + 1) = x ^ 3 + 3x ^ 2-3x + K
But it's a lot of trouble
Another method:
x^3+3x^2-3x+k
=x^2(x+1)+2x^2-3x+k
=x^2(x+1)+2x(x+1)-5x+k
Saw it
k=-5
Original formula = (x + 1) (x ^ 2 + 2x-5)

x（x-1）=56×2

X2―x=112
The solution is x = 112 or X ― 1 = 112
So X1 = 112, X2 = 113

It is known that the quadratic trinomial 2x2 + 3x-k has a factor (2x-5). Find another factor and the value of K

Let another factor be (x + b), then (2x-5) (x + b) = 2x2 + 2bx-5x-5b = 2x2 + (2b-5) x-5b = 2x2 + 3x-k, then 2B − 5 = 3 − 5B = − K, the solution is: B = 4K = 20. Then another factor is: x + 4, k = 20

2/5*X-3/5*（X-2/5*X）/（1+3/5）=56

（2/5）X－（1－2/5）X/（1＋3/5）＝56
0.4X-0.4X/1.6=56,
0.4X-0.25X=56,
0.15X=56,
X=56/0.15
=56/(3/20)
=56*20/3
=1120/3.

Square factorization of 4 (2x's square-x + 1) (x's square-2x + 3) - (3x's square-3x + 4)

4（2x²-x+1）（x²-2x+3）-（3x²-3x+4）²=4(2x^4-4x³+6x²-x³+2x²-3x+x²-2x+3)-(9x^4-9x³+12x²-9x³+9x²-12x+12x²-12x+16)=4(2x^4-5x³+9x...