Xét số hạng tổng quát:

\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)=\(\left(k^2+\frac{1}{2}\right)^2-k^2\)

= \(\left(k^2+\frac{1}{2}-k\right)\left(k^2+\frac{1}{2}+k\right)\)

Thay k từ 1 đến 12 ta được:

A=\(\frac{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(132+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(152+\frac{1}{2}\right)}\)=\(\frac{\frac{1}{2}}{152+\frac{1}{2}}=\frac{1}{305}\)

a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)

= \(\sqrt{7}+1-\sqrt{7}+1=2\)

=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)

b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

             =  \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)

= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)

=>  B=\(\sqrt{5}+1\)

c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)

=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)

                 =  \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

                =  \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)

=> A=\(\sqrt{5}\)

Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

= \(A-\sqrt{6-2\sqrt{5}}\)

= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1

Ta có : \(C=3x^2-2x+3y^2-2y+6xy-100\)

= \(\left(3x^2+3y^2+6xy\right)-\left(2x+2y\right)-100\)

=\(3\left(x+y\right)^2-2\left(x+y\right)-100\)

=\(3\cdot5^2-2\cdot5-100=-35\)

1. There(be)_will be_________ten more classrooms in our school next year.

2. Mai(be)_______had been______sick for two days.

3. I(have)______had_______to go to the dentist's last Sunday.

4. The children(play)_play_____________football in the schoolyard every Sunday morrning.

5. They(not tell)_______didn't tell_____________us about their plan last week.

6. They(build)_______built__________many bridges in the region last year.

7. Listen! I think the telephone(ring)_______is ringing_______.

8. He(be)_____was_________absent from class since last Friday.

9. We(go)______went______for a picnic to the Cuc Phuong National Park last month.

10. ''Romeo and Juliet'' written by Shakespeare is the best tragedy I(ever/see)_____have ever seen_______.

1. There(be)_____are_____ten more classrooms in our school next year.

2. Mai(be)___was__________sick for two days.

3. I(have)________had_____to go to the dentist's last Sunday.

4. The children(play)____play__________football in the schoolyard every Sunday morrning.

5. They(not tell)______didn't tell______________us about their plan last week.

6. They(build)_________built________many bridges in the region last year.

7. Listen! I think the telephone(ring)____is ringing__________.

8. He(be)_______was_______absent from class since last Friday.

9. We(go)____went________for a picnic to the Cuc Phuong National Park last month.

10. ''Romeo and Juliet'' written by Shakespeare is the best tragedy I(ever/see)____have ever seen________.

Ta có: \(x=\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}+\sqrt[3]{\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)

=> \(x^3=\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+3\cdot\sqrt[3]{\left(\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)\left(\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)}\cdot x.\)

    =  \(-b+\sqrt[3]{\frac{b^2}{4}-\left(\frac{b^2}{4}+\frac{a^3}{27}\right)}\cdot x\)

=\(-b+\sqrt[3]{\frac{a^3}{27}}\cdot x=-b+\frac{a}{27}\cdot x\)

=> \(x^3+b=\frac{a}{27}\cdot x\)

Vậy  \(x^3+ax+b=\frac{a}{27}\cdot x+ax=\frac{28a}{27}\cdot x\)