F (2x + 1) = x2-4x + 1 to find f (x) substitution method Application of substitution method Let t = 2x + 1, then X1 = (t-1) / 2, That is, f (T) = ((t-1) / 2) ^ 2 + 2 =1/4t^2-1/2t+9/4 ∴f（x）=1/4x^2-1/2x+9/4 Why didn't the last f (T) = f (x) find X1 = (t-1) / 2 before

F (2x + 1) = x2-4x + 1 to find f (x) substitution method Application of substitution method Let t = 2x + 1, then X1 = (t-1) / 2, That is, f (T) = ((t-1) / 2) ^ 2 + 2 =1/4t^2-1/2t+9/4 ∴f（x）=1/4x^2-1/2x+9/4 Why didn't the last f (T) = f (x) find X1 = (t-1) / 2 before

Because x is an arbitrary number, so t is also an arbitrary number. In fact, X and T are the same, but in different forms. Both represent arbitrary numbers

Solve the equation (2x ^ 2-1) ^ 2 + 4x ^ 2-1 = 0 by substitution method. Let y = 2x ^ 2-1, then the original equation will be transformed into a quadratic equation with one variable about y, which is: 2x ^ 2-1 is not (2x) ^ 2-1, 4x ^ 2 is not (4x) ^ 2

Let y = 2x ^ 2-1, then y ^ 2 + 2 (y + 1) - 1 = 0, namely y ^ 2 + 2Y + 1 = 0, namely (y + 1) ^ 2 = 0, then y = - 1, namely 2x ^ 2-1 = - 1, then x ^ 2 = 0, then x = 0

The solution equation: (1) x2 + 4x = 10, (2) x2-6x = 11, (3) x2-2x-4 = 0
On a rectangular ground of 35cm long and 26cm wide, two roads of the same width are built perpendicular to each other, and the remaining part is planted with flowers and plants. If the area of the remaining part is 850m2, what is the width of the road?

1. X ^ 2 + 4x = 10 (x + 2) ^ 2-4 = 10 (x + 2) ^ 2 = 14 (x + 2) = positive and negative root sign 14x = (positive and negative root sign 14) - 22, x ^ 2-6x = 11 (x-3) ^ 2-9 = 11 (x-3) ^ 2 = 20 (x-3) = positive and negative double root sign 5x = (positive and negative double root sign 5) + 33, x ^ 2-2x-4 = 0 (x-1) ^ 2-1-4 = 0 (x-1) ^ 2 = 5 (x-1) = positive and negative root sign 5x = (positive and negative root sign 5) +

There is a strange three digit number, which is exactly divided by 7 after subtracting 7, exactly divided by 8 after subtracting 8, exactly divided by 9 after subtracting 9______ ．

Because this strange three digit is divided by 7, 8 and 9, so this three digit is the least common multiple of 7, 8 and 9, 7 × 8 × 9 = 504

A three digit number can be divided by 7 after subtracting 6, by 8 after subtracting 7, and by 9 after subtracting 8. What's the number

This number can also be said to be divisible by 7 after adding one, by 8 after adding one, and by 9 after adding one
Use 7 x 8 x 9 - 1 = 503
prove:
503 - 6 = 497 = 71 x 7 divisible by 7
503 - 7 = 496 = 62 x 8 divisible by 8
503 - 8 = 495 = 55 x 9 divisible by 9
Answer: 503

A three digit number is exactly divided by 8 after subtracting 7. It is exactly divided by 9 after subtracting 8. It is exactly divided by 7 after subtracting 9. How much is this three digit number?

This number is divided by 8, 7 by 9, 8 by 7, 9-7 = 2, 7 by 8, 7 by 9, 8 by 9, then this number + 1 can be divisible by 9 and 8. This number = 72K - 1 should be divisible by 7, that is 72K - 1 = (70K + 2) + (2k - 3) obviously 2K - 3 should be divisible by 7, Kmin = 5, this number minimum = 72 * 5 - 1 = 359, because 9

A three digit number minus 6 can be divided by 7, minus 7 can be divided by 8, minus 8 can be divided by 9, what is the three digit number

Question: three digit x, X-6 |: 7, X-7 |: 8, X-8 |: 9, find x, let x = Y-1, then X-6 = Y-7, X-7 = Y-8, X-8 = Y-9, then y |: 7,8,9 is known. Then y = 7 * 8 * 9 * k = 504kx = 504k-1, easy to see x = 503 or write a congruence: x = = 6 mod 7 = = 7 Mod 8 = = 8 mod 9x + 1 = = 0 mod 7,8,9x = - 1mod 504, easy to get X

If you subtract 7 from a three digit number, you will be divisible by 7; if you subtract 8, you will be divisible by 8, and if you subtract 9, you will be divisible by 9?

7. Common multiple of 8 and 9

What is the set of even numbers divisible by 111?
Write the set and explain it

Empty set
Even numbers cannot be divided by 111

A set consisting of all even numbers divisible by 111
The answer is empty set. Why can't it be 0?

Zero is not divisor