a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B

=>B/A=1/100

b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)

\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)

\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

=>A/B=25

So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2

S=

=50/50+50/49+50/48+...+50/2

=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)

=50

P=

P=(1/49+1)+(2/48+1)+...+(48/2+1)+1

P= 50/49+50/48+....+50/2+50/50=1

vậy s/p = 1/50

3A=1.2.3+2.3.(4-1)+............+49.50.(51-48)

3A=1.2.3+2.3.4-1.2.3+.......+49.50.51-48.49.50

3A=49.50.51

\(\Rightarrow\)A=\(\frac{49.50.51}{3}=41650\)

1.2=1.2.3-0.1.2

2.3=2.3.4-1.2.3

3.4=3.4.5-2.3.4

..............

49.50=49.50.51-48.49.50

=49.50.51-0.1.2

=49.50.51

=124950

a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)

c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

=6

a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)

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

\(A=4\)

b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)

\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)

\(B=2-\sqrt{3}-1+\sqrt{3}\)

\(B=1\)

c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)

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

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

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

\(C=-4\)

d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)

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

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

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

\(D=6\)

a) \(=\frac{3}{17}-\frac{2}{345}+\frac{5}{12}+\frac{2}{345}-\frac{3}{17}+\frac{1}{12}\)

\(=\left(\frac{3}{17}-\frac{3}{17}\right)+\left(\frac{2}{345}-\frac{2}{345}\right)+\left(\frac{5}{12}+\frac{1}{12}\right)\)

\(=\frac{6}{12}=\frac{1}{2}\)

b) \(=-\frac{49}{25}-\frac{5}{36}+\frac{4}{123}+\frac{49}{25}-\frac{4}{123}-\frac{1}{36}\)

\(=\left(-\frac{49}{25}+\frac{49}{25}\right)+\left(\frac{4}{123}-\frac{4}{123}\right)-\left(\frac{5}{36}+\frac{1}{36}\right)\)

\(=-\frac{6}{36}=-\frac{1}{6}\)

Cảm ơn