Seeking 100 mathematics problems in Volume 1 of grade 2 of junior high school 50 factorization 50 channel integral multiplication and division

Seeking 100 mathematics problems in Volume 1 of grade 2 of junior high school 50 factorization 50 channel integral multiplication and division

1. (2a + 3b) (a-2b) - (3a = 2b) (2b-a) 2.4m square + 8m + 43. (x square + 4) square + 8x (x square + 4) + 16x Square) 4

Find the right number for the following letters. (a, B, C, D stand for different numbers)
a b c d a=（ ） b=（ ）
c=（ ） d=（ ）
× 4
___________
d c b a

Because D is one digit, the result is four digits
The result of a × 4 must be one digit
A = 1 or a = 2
If a = 1, then a bit must be equal to 1
But the number of bits from D × 4 must be 1
1×4=4
2×4=8
……
All the way up to 9, you can find that none of the seats is 1!
So a can't be equal to 1!
A is equal to 2!
Since a = 2, the bits of D × 4 must be 2
D may be equal to 3.8
Now let's look at the highest position. The highest position is d
So let's assume that d = 3. From the above, we can know that a = 2 is certain
So is it possible for 2 × 4 to be 3? -- no way!
So d = 8!
Through the derivation of D, it is impossible for B × 4 to enter 1!
Then B may be equal to 1 or 2
Let's deduce it from ten digits
Suppose B = 2 (note that we are going into 3). Therefore, C × 4 is true only when the number of bits is 9
1×4=4
2×4=8
……
No! Therefore, B cannot be equal to 2
So B = 1!
2 1 c 8 a=2 b=1
c=（ ） d=8
× 4
___________
8 c 1 2
There's C left
At this time, we should enter the number (3), so we should not enter the number (3)
1×4=4
2×4=8
……
C may be equal to 2 or 7
Now try one by one
Put 2 in first. The result can be imagined, it is wrong!
So C = 7
2178 * 4 = 8712 ha ha

Help to do some math problems in volume one of junior three
To solve the equation: (1) the square of X -- 3x = 4 (2) the square of X + 6x -- 11 = 0
2. It is known that one root of the square of X + 3x + M = 0 is 4. Find its other root and the value of M
3. The quadratic equation of one variable is known: the square of X -- 2x + M = 0
(1) If the equation has two real roots, find the value range of M; (2) if the two real roots of the equation are x1, X2, and X1 + x2 = (x1 multiplied by x2), find the value of M

(1) the square of X -- 3x = 4x ^ 2-3x-4 = 0 (x-4) (x + 1) = 0x = 4x = - 1 (2) the square of X + 6x -- 11 = 0 is solved by formula method, and x = - 6 + √ 6 ^ 2 + 4 * 11 / 2 = - 3 + √ 20 or x = - 6 - √ 6 ^ 2 + 4 * 11 / 2 = - 3 - √ 202 is obtained