Ask for an explanation of the math problem in the second year of junior high school, in a hurry!

Because 1/A + 1/B = 1/A + 1/B

So (a+b) 2=ab, a 2+b 2=-ab

A part b + b part a = ) a 2 + b 2) ab = -1

Known: 1 a+1 b=1 (a+b); (a+b) ab=1 (a+b); i.e. (a+b)(a+b)=ab; Arrange a 2 + b 2 = -ab

b/a+a/b=(b^2+a^2)/ab;Bringing in what is known, we get -1

Grass....I suspect it's wrong, you see, I solved it.

1/a+1/b=b/ab+a/ab=(a+b)/ab

Therefore, it is known that (a+b) ab=1 (a+b) is then simplified to obtain (a+b).

1. You can break down the conditional formula to get square A plus square b = ab

Negative 1 has a known square of a + square b = negative ab

1. Connecting the OB, it can be proved that the OHBC four points are circular (angle BHC = angle BOC = 90). Thus the earthly ant can prove ohf= acb=45.

2. ob 2 = bh 2 + oh 2 + 2 * bh * oh * cos45 (buried 1 2) bc 2

1 2) (bh 2 + ch 2).

This will solve it.

: root number 3:2 (because the right-angled side of thirty degrees is equal to half of the hypotenuse).

kmh (how can an airplane fly so slow.) The person who asks the question does not have a long head), 30002 + 40002 = 50002

3. Root number (1502 + 802) = root number 28900 (open the root number by yourself).

Prove: 1) The straight line L does not intersect the bottom edge AB, and the straight line L is parallel to the bottom edge AB, so the angle CBA=angle BCF=45 degrees, and BF is perpendicular to the straight line L, so we can get: the triangle AEC is an isosceles right triangle, so: BF=CF

In the same way, it can be obtained that the triangle bfc is also an isosceles right triangle, so: ae=ce, so: ef=ce+cf=ae+bf

From the question: the angle FDB and the angle ADE are the opposite top angles, which are equal.

In the RT triangle angle DEA, ed=ae*ctg(angle ADE), in the RT triangle angle DFB, FD=BF*CTG(angle FDB)=BF*CTG(angle ADE).

So: EF=ED+DF=AE*CTG(Angular ADE)+BF*CTG(Angular ADE).

AE+BF)*CTG (Angle ADE).

3x-2k=4(x-k)+1

Simplified to x-2k+1=0 x=2k-1

x>0 gives 2k-1>0

Answer: k> is a positive number.

3x-2k=4(x-k)+1

3x-2k=4x-4k+1

x=2k-1

2k-1＞0

2k＞1k＞1/2

So, when k 1 2, the solution of the equation is positive.

Express x in k.

That is, x=2k-1

Obviously, when 2k-1>0 x is positive.

So solving this inequality yields k >

When k >, the solution of the equation 3 x 2k = 4 (x k) + 1 is a positive number, I hope it can help you.

3x–2k=4(x–k)+1

3x–2k=4x–4k+1

x=4k-1-2k=2k-1

So at 2k-1>0 x is a positive number.

Hence k>1 2

3x-2k=4x-4k+1

3x-4x=2k-4k+1

x=-2k+1

x=2k-1

So when k>1 2, the solution of the equation is positive.

Untie; From the meaning of the title; 3x 2k=4(x k)+1 greater than 0 3x 2k 0 ; 4(x k)+1 0 solution { by ; } Deed; x plus a quarter k}

k 1/2; k One quarter.

k One quarter.

3x-2k=4x-4k+1,3x-4x=-4x+2x+1,-x=-2x+1,x=2x-1,and because it's positive, it's x 0, so 2x-1 0, so 2x 1 x 1 2

This kind of question? Friend, if you don't, you'll have to reflect! You have to master the method, it's all basic!

The process is as follows....

May it help you.

3x-2y=m-4---1）

x+3y=m+5---2）

3x+2y=5m-8---3）

3)-(1), obtain: 4y=4m-4, finishing: y=m-1---4)(2) 3-(1), obtain: 11y=2m+19,--5).

Substituting (4) into (5) obtains: 11m-11=2m+19, finishing: 9m=30, m=10 3

Hope it helps