1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)+11+(-12)

=(1+3+5+7+9+11)+[(-2)+(-4)+(-6)+(-8)+(-10)+(-12)]

= 36+-42

=-6

(-1)+2+(-3)+4+(-5)+6+(-7)+8+(-9)+10+(-11)+12

=[(-1)+(-3)+(-5)+(-7)+(-9)+(-11)]+(2+4+6+8+10+12)

=(-36)+42

=6

làm đúng có tick các bạn ơi

So sánh: mk làm luôn nè:

Ta có: \(\frac{10}{11}>\frac{10}{11+12};\frac{11}{12}>\frac{11}{11+12}\)

\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10}{11+12}+\frac{11}{11+12}\)

\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10+11}{11+12}\)

MK KO BIẾT ĐÚNG KO NỮA NÊN BN CÓ THỂ THAM KHẢO CỦA CÁC BẠN KHÁC NHÉ.!!

CHÚC BẠN HỌC TỐT. ^_^

 1/2+1/2^2+1/3^2+1/4^2+.......+1/100^2<1

= 1/2 + 1/4 + 1/9 + ... + 1/10000

có : 100 - 1 + 1 = 100 số hạng 

1 = 1/100 + 1/100 + ... + 1/100

suy ra  1/2+1/2^2+1/3^2+1/4^2+.......+1/100^2<1

T làm biếng lắm; làm C thôi

\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)

Làm tương tự ta được A > 1/15

câu a

\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)

\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)

@Ngô Tấn Đạt

Câu 2 sai đề, thử rồi

ta có:1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
...
1/100^2<1/99.100=1/99-1/100
=> A=1/2^2+1/3^2+1/4^2+.....+1/100^2<1/4+1/2-1/3+1/3-1/4+...+1/99-1/100
<1/4+1/2-1/100<1/2

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}\)

\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)

\(< \frac{1}{2}-\frac{1}{100}\)

\(< \frac{1}{2}\)

a,1/10²+1/11²+1/12²+...+1/100²<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100

                                                                                                    =1/9-1/100=91/900<3/4

Vậy 1/10²+1/11²+1/12²+...+1/100²<3/4

b,1/2²+1/3²+1/4²+...+1/100²<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

                                                                                        =1-1/100=99/100

Vậy 1/2²+1/3²+1/4²+...+1/100²<99/100

c,1/2²+1/3²+1/4²+...+1/100²<1/2²+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)

                                                                                       =1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100

Vậy 1/2²+1/3²+1/4²+...+1/100²<3/4

Ta có :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....................+\dfrac{1}{100^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{99.100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}< 1\)

\(\Rightarrow A< 1\) \(\rightarrowđpcm\)

Ta có

\(\dfrac{1}{2^2}< \dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(.........\)

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

Cộng theo vế ta có:

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< 1-\dfrac{1}{100}< 1\)

Vậy \(A< 1\left(dpcm\right)\)

\(A=\dfrac{1}{2^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}< 1\)