As shown in figure 11.1-7, to know that a, D and E are on the same straight line, and △ bad ≡ ace Preferably before 15; 30

As shown in figure 11.1-7, to know that a, D and E are on the same straight line, and △ bad ≡ ace Preferably before 15; 30

1, because they are congruent, so BD = AE, ad = CE, so BD = de + CE
2. If parallel, then BDE = AEC (equal internal stagger angle), because congruent, so BDA = AEC, so BDA = BDE
So BDA is a right angle, so abd is a right triangle

As shown in the figure, point P is a point in △ ABC. Try to judge the size relationship between ab + AC and Pb + PC, and explain the reason
Mathematics Evaluation Handbook, page 21, question 5

AB+AC>PB+PC
Reason: because: extend BP AC to d.ab + ad > BD = Pb + PD
Because: PD + CD > PC two formula add
So: ab + AD + PD + CD > Pb + PD + PC sell to PD
So: ab + (AD + CD) > Pb + PC
So: ab + AC > Pb + PC

(1) Given that n is a rational number, find the minimum value of quadratic trinomial n ^ 2-4n + 5

n²-4n+5=(n-2)²+1
When n = 2, the minimum value of the formula is 1

① The minimum value of (A-2) &# 178; + 4 is the maximum value of (A-3) &# 178;)

① The minimum value of (A-2) &# + 4 is 4
② The maximum value of 8 - (A-3) & # is 8

CN = (4n-5) * (1 / 2) ^ (n-1), find the maximum value of CN

f(x)=(4x-5)/2^(x-1)
f'(x)=[4*2^(x-1)-(4x-5)*2^(x-1)*ln2]/[2^(x-1)]²
=[4-(4x-5)ln2]/2^(x-1)=0
4x-5=4/ln2
X = (4 / LN2 + 5) / 4 = 1 / LN2 + 5 / 4 is about 2.7
Here n > = 1
So x > = 1
So 1

If (2 + √ 2) &# 178; = a + B √ 2 (a, B are rational numbers), then a + B=

=4+2+4√2=a+b√2
a+B=10

If the rational number a > b, then the relationship between a and B is the same

If the rational number a > b, then the size relationship between a and B is unable to judge, because AB does not know whether it is greater than 0

If (1 + √ 2) &# 178; = a + B √ 2 (a, B are rational numbers), then a + B=

If (1 + √ 2) &# 178; = a + B √ 2
Then a + B √ 2 = (1 + √ 2) &# 178; = 1 + 2 √ 2 + 2 = 3 + 2 √ 2
So a = 3, B = 2
So a + B = 5

If rational numbers a and B satisfy A-2 + (1-B) &# 178; = 0?
Try to find 1 / AB + 1 / (a + 1) (B + 1) + 1 / (a + 2) (B + 2) + +(hint: 1 / 2 × 3 = 1 / 2-1 / 3, 1 / 3 × 4 = 1 / 3-1 / 4 ^)

If rational numbers a and B satisfy A-2 + (1-B) & # = 0, then A-2 = 0, 1-B = 0A = 2, B = 11 / AB + 1 / (a + 1) (B + 1) + 1 / (a + 2) (B + 2) + +1/(a+2010)(b+2010)=1/2*1+1/3*2+1/3*4+/14*5+++++++.+1/2012*2011=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.-1/2011+1/...

If the rational numbers a and B satisfy | a + B | + (B-4) & # 178; = 0, then the value of a + B is

|a+b|+（b-4）²=0
Because | a + B | > = 0, (B-4) & # 178; > = 0
The equation holds only when | a + B | = 0, (B-4) & # 178; = 0
That is, a + B = 0