Junior 2 math problems, fast forward, fast forward 2nd junior high school math problems

1: Obviously: a=2-b Generation:

2-b) square + b square + 4b

4-4b + b square + square + 4b

4+2b squared.

No answer! Because when a=0, b=2, the result is 8, and when a=1, b=1, the result is 6

The original question is changed to the quadratic of a - the quadratic of b +4b to have a unique answer The result is 42: (1) = 8 (m+n) square + 4mn No solution!

2) I don't understand!

1. From the known a=2-b, so a 2+b 2+4b=(2-b) 2+b 2+4b=4+b 2-4b+b 2+4b=2b 2+4

2 (1), original formula = 9 (m 2 + n 2 + 2mn) - (m 2 + n 2-2mn) = 18 mn + 2mn = 20mn

2(2), original formula = a (n-1) (a 2-2ab + b 2) = a (n-1) (a-b) 2

1.Because the square of a plus the square of b plus 4ab

So get the square of the parentheses a + b back brackets.

So the answer is 4

9 (m+n) to the second (m-n) to the second (m-n).

3(m+n)]^2-(m-n)^2

3m+3n)+(m-n)][3m+3n)-(m-n)](4m+2n)(2m+4n)

4(2m+n)(m+2n)

1.It is known that a+b=2 is found to be the quadratic of a +b +4ba 2+b 2+4b

a^2+2abb^2+4b-2ab

a+b)^2+4b-2ab

a+b)^2+2b(2-a)

4+2b^2

y=180 divided by x

y=60min, substitution to get x=3, so the value range is 0 to 3, note that it is not equal to 0

1.（b-c）2+（2a+b）（c-b）=0

b-c）²-b-c）（2a+b）=0

b-c）（b-c-2a-b）=0

b-c）（c+2a）=0

b-c=0 or c+2a=0

b = c or c = -2a (rounded).

So ABC is an isosceles triangle.

1/2（2a2+2b2+2c2-2ab-2bc-2ac）

1/2（a²-2ab+b²+a²-2ac+c²+b²-2bc+c²）

1/2[（a-b）²+a-c）²+b-c）²]

1/2[-(1/20)x+20-(1/20)x-19]²+1/2[(1/20)x+21-(1/20)x-19]²+1/2[-(1/20)x+20-(1/20)x-21]

1/2[1-（1/10）x]²+1/2*4+1/2[1+（1/10）x]²

1/2[1-1/5x+1/100x²+4+1+1/5x+1/100x²]

1/2（6+2/100x²）

3+x²/100

3.（a+b）^1=a+b

a+b）^2=a²+2ab+b²

a+b）^3=a^3+3a²b+3ab²+b^3

a+b）^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5

n2=m+2

m²-n²=（m-n）（m+n）=n+2-m-2=n-m

So m+n=-1

m3-2mn+n3

m*m²-2mn+n*n²

m（n+2）-2mn+n（m+2）

mn+2m-2mn+mn+2n

2m+2n=-2

x +2xy+y -a(x+y)+16=(x+y) -a(x+y)+16, i.e. (x+y) -a(x+y)+16 is a perfectly squared test, so x+y=t

So t -at+16=t -at+4 , so 2 1 4=|-a|

So a = 8 or -8

x+y)^2-a(x+y)+4^2

Let x+y=q

q^2-aq+4^2

It's easy.

q+4) 2 or (q-4) 2

Count yourself as A

xx+2xy+yy-a(x+y)+16=(x+y)(x+y)-a(x+y)+16

Because it's completely flat.

So =(x+y+4)(x+y+4)=(x+y)(x+y)-a(x+y)+16

So a=8

And a= -8 also holds.

So a=8

This is also written by myself, I should be right, I hope it can help you, I'm also a sophomore in junior high school (*

If you think about it this way, if x+y is an unknown variable b, i.e., let x+y=b, then your problem becomes that if b 2+ab+16 is a perfect flat method, and the formula of the perfect flat method is a 2+2ab+b 2 or a 2-2ab+b 2, you can find a=b and b=4 substitution is not difficult to calculate a=8 or -8

Original = (x+y) -a(x+y)+4

Think of (x+y) as a whole, because it's perfectly flat.

So 2 1 4 = a

So a=8

1. Let both sides be x y. According to the Pythagorean theorem: x 2 + y 2 = 10000 x y = 4800

How to solve this equation, I thought of a simple way, (x+y) 2=x 2+y 2+2xy, you know (x+y) 2=19600, and then the root number, calculate x+y, the perimeter is 2(x+y), bring it in and calculate, I say this, can you understand.

2. x 2+y 2=2500 x+y+50=120, the same is true for solving this problem, but replacing x+y with xy, I think the reason why you are asking is because you don't know how to solve this equation.

It's so simple that you ask. Pythagorean theorem has no ?。。

1. Let both sides be x y. According to the Pythagorean theorem: x 2 + y 2 = 10000 x y = 4800

Solve the equation and you're good to go.

2: Hook 3 strands 4 strings 5. You'll know the answer at a glance.

According to the Pythagorean theorem, the equation is solved again according to the Pythagorean theorem x 2+y 2=2500 x+y+50=120.

This is a one-time function, let y=kx+b substitute two points (,64)(2,60) to get k=-20 b=100, so y=-20x+100

The inverse proportional relationship decreases because x increases and y decreases.