NHỚ PHẢI TÍCH TỚ ĐẤY

S=1-1/2-1/3+....+1/29-1/30

=1-1/30

=29/30

S = 1/1x1/2+1/2x1/3+1/3x1/4+...+1/28x1/29+1/29+1/30

S = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/28-1/29+1/29+1/30

Đến đây ta triệt tiêu,còn lại:

S = 1/1-1/30

S = 29/30

Mình chắc chắn lun!

Có lẽ bạn viết lộn đề , dấu cuối phải là dấu nhân mới đúng/

S= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{29}-\frac{1}{30}\)

S=\(1-\frac{1}{30}\)

S=\(\frac{29}{30}\)

AI KẾT BN KO!

TIỆN THỂ TK MÌNH LUÔN NHA!

KONOSUBA!!!

A=\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{999.1000}\)\

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{999}-\frac{1}{1000}\)

\(A=\frac{1}{1000}-1\)

A=\(\frac{999}{1000}\)

giải giùm e vs ạ

999/1000

1/1.2+1/2.3+1/3.4+1/4.5+.................+1/9990999.9991000

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.................+1/9990999-1/9991000

=1-1/9991000

=9990999/9991000

chế vừa thôi cụ

1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)

2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)

3:

\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)

4:

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

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

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

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

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

\(=2\sqrt{2}+\sqrt{5}+1\)

2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)

3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)

\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)

\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)

\(=-\sqrt{a}\)

4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

help

Bài 2:

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{999\cdot1000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

=1-1/1000

=999/1000

a: =>78/4+x=30

=>x=30-78/4=42/4=10,5

b: =>x-2/3=1/6

=>x=1/6+2/3=5/6

c: =>16/3-x=7/4

=>x=16/3-7/4=43/12