1/2*4+1/4*6+1/6*8+1/8*10+...+1/98*100

=1/2*2(1/2*4+1/4*6+1/6*8+1/8*10+...+1/98*100)

=1/2(2/2*4+2/4*6+2/6*8+2/8*10+...2/98*100)

=1/2(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+...+1/98-1/100)

=1/2[(1/2-1/100)+(1/4-1/4)+(1/6-1/6)+...+(1/98-1/98)

=1/2*(1/2-1/100)=1/2*(50/100-1/100)=1/2*49/100=49/200

2. Our class( start) starts at 7.15 and( finish) finishes at 11.45 every morning

3. She gets fat because she is always tasting things while she( cook) cooks

4. Bill decided( buy) to buy a new car rather than a used one

5. Long( fall) fell down the stairs this morning and ( break) broke his leg

6. George is interested in( take) taking an art class

7. The package has( be) to be there tomorrow. Will it( be) be there in time?

- Don’t worry. I am going to send it by express mail.

1. Did you (see) see the film on television last night?

2. Our class( start) starts at 7.15 and( finish) finishes at 11.45 every morning

3. She gets fat because she is always tasting things while she( cook) cooks

4. Bill decided( buy) to buy a new car rather than a used one

5. Long( fall) fell down the stairs this morning and ( break) broke his leg

6. George is interested in( take) taking an art class

7. The package has( be) to be there tomorrow. Will it( be) 0 there in time?

- Don’t worry. I am going to send it by express mail.

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)

=>3a=b+c+d

    3b=a+c+d

    3c=a+b+d

    3d=a+b+c

=>a=b=c=d

=>\(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

CM + CD + D + C + M + XC = CMCDDCMXC = 2990

a)

\(=a^2\left[\left(b-a\right)+\left(a-c\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=-a^2\left(\left[\left(a-b\right)+\left(c-a\right)\right]\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=-a^2\left(a-b\right)-a^2\left(a-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(-a^2+c^2\right)+\left(a-c\right)\left(-a^2+b^2\right)\)

\(=\left(a-b\right)\left(a+c\right)\left(c-a\right)+\left(c-a\right)\left(-a^2+b^2\right)\)

\(=\left(c-a\right)\left[\left(a-b\right)\left(a+b\right)-a^2+b^2\right]\)

\(=\left(c-a\right)\left[ac+a^2-bc-ba-a^2+b^2\right]\)

\(=\left(c-a\right)\left(ac-bc-ba+b^2\right)\)

b)

\(=\left(1-2a+a^2\right)-\left(c^2-2cb+b^2\right)\)

\(=\left(1-a\right)^2-\left(c-b\right)^2=\left[a-1+\left(c-b\right)\right]\left[a-1-\left(c-b\right)\right]\)

\(=\left[\left(1+a\right)-\left(c-b\right)\right]\left[\left(1+a\right)+\left(c+b\right)\right]\)

\(=\left(1+a-c+b\right)\left(1+a+b+c\right)\)

\(A=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.......+\frac{1}{101}-\frac{1}{400}\right)\)

\(A=\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\right]\)

=>đpcm

a)

Tổng 17 số đầu tiên là

(6x1-3)+(6x2-3)+....+(6x17-3)

=6(1+2+3+...+17)-3x17

=6x153-17

=867

b)

Tích 100 số hạng bất kì là

\(\left(6m-3\right)\left[6\left(m+1\right)-3\right].......\left[6\left(\left(m+99\right)-3\right)\right]\)

\(=3\left(2m-1\right)3\left[2\left(m+1\right)+1\right]......3\left[2\left(m+99\right)+1\right]\)

\(=3^{100}\left(2m-1\right)\left[2\left(m+1\right)-1\right].......\left[2\left(m+99\right)-1\right]\)

chia hết cho 3⁹⁹

Vậy tích 100 số bất kì của dãy chia hết cho 3⁹⁹