Solve the 18th problem and leave the rest alone, and solve the 18th problem?

Hello, these are homogeneous formulas, and the numerator and denominator of the conditional formula are divided by cosa (tan a +1) (tan a -1) = 2, so that tana = 3

The first question, in the same way, is the same as dividing by cosa formation (3tana-1) (2tan+3)=8 9

The second question, the hypothetical denominator is 1 = sin square + cos square, the same reason, the mouth calculated 13 10, you calculate it again.

sinα+cosα)/(sinα-cosα)=2

sinα+cosα=2(sinα-cosα)

sinα+cosα=2sinα-2cosα

So sin = 3cos

1）（3*3cosα-cosα）/（2*3cosα+3cosα）=8/9

2) sin +cos sin -cos =2 The numerator and denominator on the left side of the equal sign are divided by cosa to get tana=3

sina 2-2sin cos +1=(sina 2-2sin cos +sin +cos) (sin +cos ) This step is a transformation of 1, note this little trick] The numerator denominator is the same as dividing by cos.

tan²a-2tana+tan²a+1)/(tan²a+1)=13/10

Solve a,b coordinates (2,4),(4,2) and use the point to straight line distance formula to find the distance from a to y= x, which is the height of the triangle 6 5 5, and the ob distance 2 5, so the area of the triangle is 6

Students can do it with fixed points!

You can find the range of the inverse function:

y(4x+2)=5x-1

4xy + 2y - 5x=-1

x(4y - 5)=-1 - 2y

x=(-1 - 2y)/(4y - 5)∵4y - 5≠0

y≠5 4 is the range of the function.

hclo4 , naoh, al(oh)3,oh- +al(oh)3=alo2- +2h2ona, f

N2 +3H2 = reversible number (high pressure catalyst) = 2NH3F2 + 2Na = 2NaF

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(0,13 3), symmetrical point b with respect to the x-axis to obtain b, then the intersection point of a,b line and x-axis is the point sought.