Very simple algebra problems, help me

There is no easy way to cube, you can only try it little by little like the dichotomy mentioned upstairs, and you still have to use a calculator in the back, otherwise the amount of calculation is too large.

There is an easy way to square it.

As you said 4421

1.Start with the right, draw every two digits, and divide the number into two digits and two digits, 4421 is divided into 44'21

Open the square of 44 first, and get 6*6<44<7*7

2.The first digit is 6, 44-36 = 8, leaving the remainder and the last two digits as 821

Then multiply 6 by 20 for 120

6 remainder 101, 6 as the single digit of 120, 120 6 126, then you need to check whether 126 6 is less than 821, here is less than or equal to, so remember 126, (if it is greater than here, 6 should be subtracted by 1, remember 125.) ）

821 126 = 6 remainder 65, the second place is 66, at this time the integer part has been opened, 66 is obtained, the integer part of the square root is 66, and the back is a small tree.

4.66 20 gives 1320, and the remainder 65 with two decimal places in front of it is 6500

6500 1320 gets 4, the same check 1324 * 4<6500, established, the first decimal point is 4, get.

6500 1324 is 4 surplus 1204

Then the remainder continues to take the last two zeros to get 120400, remove the decimal point and multiply by 20 to get 664 20 13280, the following practice is the same as above. You can keep it going.

Dichotomy is okay.

For example, 9 squares, so that x 3 = 9

x=2,2^3=8;

x=3,3 3=27,then x=,<9; So x=, it seems to be less than 9, so x= is too lazy to take the calculator

In short, it is gradually approaching.

Calculator results 9 of 1 to the 3rd power

Approximation f(x) f(x0)+[f(x0)]."(x-x0) （"denotes a derivative).

i.e. 9 to the 1 3rd power 8 to the 1 3rd power + 1 3 times 1 4 = 25 12 (

Calculator results 4421 open squared

Approximation 66 squared = 4356 67 squared = 4489 take x0= squared=

4421 kelv squared Kai square - (multiplied by kai squared).

Because 6x-1 is the most segment key hail big side, there is.

Obviously 4x+6>2x+1, don't think about it.

6x-1>4x+6 gives x>7 2

2x+1+4x+6=6x+7<6x-1 is also not considered.

Consider again that (2x+1) 2+(4x+6) 2=(6x-1) 2 is a right triangle.

The square of the largest side > the flat and bright first square of the other two sides are obtuse triangles, so there is:

2x+1) 2+(4x+6) 2<(6x-1) 2 to get x>

To sum up, at the > of x, there are obtuse angles. Hold the sails.

If you haven't learned the cosine theorem.

Ideas: 1. Consider forming a triangle (the solution upstairs should also consider this modulus first). The sum of either side is greater than the third side.

2 There is a square sum of the two sides that is less than the square of the third side.

3 Eyes on the intersection of 1 and 2 in the range.

Reason for x>: 6x-1> 2x+1 and 6x-1>4x+6 and 4x+6+2x+1>6x-1 and (4x+6)-(2x+1)<6x-1

and 4x+6>0 and 2x+1>0 and 6x-1>0 and (4x+6)(4x+6)+(2x+1)(2x+1)<(6x-1)(6x-1).

Get x> x> x> x> x> x>1 6 only know that there is x> to meet all the above conditions.

Using the cosine orange annihilation theorem a square = b square + c square -2bc (cosa), cosa is the cosine value of an inner angle of the triangle, using the above formula to express this trigonometric function, and because it is an obtuse angle of the fiber code, so this value is less than which 0 is destroyed, and in the triangle, so it is not equal to -1, according to the above conditions, the inequality is listed, and the solution set is obtained.

The original is (x+1) 2+4(y 2+1)-1 and has a minimum value of 3 when x=-1 and y=0

Use mathematical induction.

When n=2, (2n) (n -1)=4*3=12, which is a multiple of 12, is true; Let the banquet be (2n+1) (2n +2n)=5*12=60, which is a multiple of 12, and is established.

Let n=k, (2k) (k -1) and (2k+1) (2k +2k) are multiples of 12.

1) When n=k+1, 2(k+1)]*k+1) 2-1]=(2k+2)[(k+1+1)(k+1-1)]=2k+2)[k(k+2)]=2k(k+1)(k+2)=2k(k+1)(k-1+3).

2k(k+1)(k-1)+6k(k+1)

2k(k^2-1)+6k(k+1)

According to the assumption, (2k) (k -1) is divisible by 12, and one of the two adjacent positive integers must be even, so 6k(k+1) is divisible by 12.

So when n=k+1, [2(k+1)]*k+1) 2-1] is divisible by Tanxun12.

So for n>1, there is (2n) (n -1) which is a multiple of 12 true.

2) When n=k+1, 2(k+1)+1][2(k+1) 2+2(k+1)]=2k+3)*2(k+1)(k+2)=4k(k+1)(k+2)+6(k+1)(k+2).

From (1), it can be seen that 2k(k+1)(k+2) is divisible by 12 (which is one step of the simplification of [2(k+1)]*k+1) 2-1], and 6(k+1)(k+2) is divisible by 12, and 2(k+1)+1][2(k+1) 2+2(k+1)] is divisible by 12.

So for n>1, there is (2n+1) (2n +2n) which is a multiple of 12 true.

2n) (n -1) = 2 (n-1) n (n 1) n-1), n, (n 1) is a continuous natural number, one of which must be a multiple of 3, at least one even number, so (2n) (n -1) = 2 (n -1) n (n state 1) is a multiple of 12.

It can be proved in a similar way.

2n+1) (2n +2n)=2 n (n 1) (2n 1) is a multiple of 12.

A + B +C -3ABC = (A + B + C - Ab-AC-BC) = (A + B +C -Ab-AC-BC) = (A + B + C) [(A + B + C) -3 (Ab + BC + CA)].

Bring in the conditions to get the pulse.

Let's do fractional addition and subtraction.

2/5x-4/3x²

The common denominator is 15x

Original = (2 3x) 15x -(4 5) 15x = (6x-20) 15x

Root number xy root number y-2 times root number x-2 = 0

xy √y-2√x-2=0

y(√x 1)-2(√x 1)=0

y-2)(√x 1)=0

y-2=0,y=4

And x>=0 will do.