How to solve the equation 2x ^ 2-3x + 1 = 0

How to solve the equation 2x ^ 2-3x + 1 = 0

2x²-3x+1=0
（2x-1）（x-1）=0
2x-1 = 0 or X-1 = 0
x1=1/2
x2=1

Calculation questions: 1.12 √ 2 / 3-6 √ 3 / 2 - & # 189; √ 24 solving binary linear equation: 1. {3x + 2Y = 5,2x-y = 8
2. {3 / x-4 / y = 1,3x-4y = 2

(1) 3x+2y=5 ① ①+②×2：7x=21
2x-y=8 ② x=3 y= - 2
(2)1/3x-1/4y=1 →4x-3y=12 ① ①×3 - ②×4：7y=28
3x-4y=2 ② y=4 x=6

LIM (5N + 4) an = 5, find the direct solutions of Liman and limn * an

Liman = 0, otherwise let Liman = a (not 0), LIM (5N + 4) an = LIM (5N + 4) Liman = infinity
Since Liman = 0, LIM (5N + 4) an = lim5n * an + lim4 * an = 5limn * an
So limn * an = 1 / 5 * LIM (5N + 4) an = 1

55555. (a total of 2000 5) divide by 13 to get the remainder

7×11×13=1001
Remember this, many topics can be used
111×1001=111111
Six consecutive ones, divisible by 13
So six consecutive 5S can also be divided by 13
2000 △ 6 = more than 333 2
So 2000 5 divided by 13 is the same as 55 divided by 13
55 △ 13 = 4,3
The remainder is 3

Simple operation: 1 / 1 × 2 × 3 + 1 / 3 × 5 × 7 + 1 / 5 × 7 × 9 + ＋1/11×13×15

1/1×3×5+1/3×5×7+1／5×7×9＋…… ＋1/11×13×15
=(1/1×3-1/3×5+1/3×5-1/5×7+1／5×7-1/7×9＋…… ＋1/11×13-1/13×15)÷4
=(1/1×3-1/13×15)÷4
=(1/3-1/195)÷4
=64/195÷4
=16/195

Finding the general solution or special solution of (x ^ 2 + y ^ 2) DX xydy = 0 differential equation

This is a first order homogeneous differential equation
(x^2+y^2)dx-xydy=0
dy/dx=(x²+y²)/(xy)
dy/dx=((x/y)²+1)/(x/y)
Let u = Y / X
Then dy = Du * x + DX * u
dy/dx=(du/dx)*x+u
arcsinx-x
(du/dx)*x+u=(u²+1)/u=u+1/u
du/dx=1/(xu)
u*du=dx/x
Two sides integral
(1/2)u²=lnx+C
Substituting u = Y / X back
(1/2)(y/x)²=(lnx)+C
y²=2x²((lnx)+C)
This is the general solution of the differential equation~

Given that two straight lines y = AX-2 and y = (a + 2) x + 1 are perpendicular to each other, then a is equal to ()
Given that two straight lines y = AX-2 and y = (a + 2) x + 1 are perpendicular to each other, then a is equal to ()
Who can write more detailed steps
Write down any formula

The two lines y = AX-2 and y = (a + 2) x + 1 are perpendicular to each other,
It shows that the product of X coefficient of two linear equations is - 1
That is: a (a + 2) = - 1
(a+1)(a+1)=0
a=-1

351 △ 58 vertical

Find LIM (x →∞) under the root sign (x ^ 2 + x) - under the root sign (x ^ 2-x), thank you~

(x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x ^ 2) + √ (1-1 / x ^ 2)]

(5-7X)/8=(7-5X)/7-1

(5-7x) / 8 = (7-5x) / 7-1 multiplied by 56 7 * (5-7x) = 8 * (7-5x) - 56 35-49x = 56-40x-56 35 = 9x x x = 35 / 9