\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{20}{41}\)

Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{10}{31}\)

Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...

\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)

\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)

Mà \(n\ne\dfrac{2}{3};n\in Z.\)

\(\Rightarrow n\in\left\{1;-1\right\}.\)

Ta có: \(\frac{1}{5^2}<\frac{1}{4\cdot6}\)

\(\frac{1}{7^2}<\frac{1}{6\cdot8}\)

...

\(\frac{1}{103^2}<\frac{1}{102\cdot104}\)

Do đó: \(\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{103^2}<\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\cdots+\frac{1}{102\cdot104}\)

=>\(S<\frac12\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\cdots+\frac{2}{102\cdot104}\right)\)

=>\(S<\frac12\left(\frac14-\frac16+\frac16-\frac18+\cdots+\frac{1}{102}-\frac{1}{104}\right)\)

=>\(S<\frac12\left(\frac14-\frac{1}{104}\right)=\frac12\cdot\frac{26-1}{104}=\frac12\cdot\frac{25}{104}\)

=>\(S<\frac{25}{208}<\frac{25}{160}\)

=>S<5/32

đk : \(n\ne-\dfrac{1}{3}\)

gọi d là ƯCLN(18n+3,21n+7)

ta có 18n+3chia hết cho d

          21n+7 chia hết cho d

⇔21n+7-18n-3 chia hết cho d

⇔126n+42-126n-21 chia hết cho d 

21 chia hết cho d

⇒d∈Ư(21)=1;3;7;21

n ≠ 3k-1;3k-3;3k-7;3k-21

Bài 2: 

Gọi tử là x

Mẫu là x+6

Theo đề, ta có:

\(\dfrac{x+3}{x+5}=\dfrac{4}{5}\)

=>5x+15=4x+20

=>x=5

(3) \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

\(\Leftrightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\left(luôn đúng\right)\)

(4)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a +b+c}\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)

\(\Leftrightarrow1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+1+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{b}{c}+1\ge9\)

\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge3+2+2+2\ge9\) (đpcm)

=57.(157-98-57):57 (bn đặt thừa số chung nhé)

=157-98-57=2 (sai thôi nhé mk ko bấm máy)

=\(=\frac{57\left(157-98-57\right)}{57}=2\)

x-3/-9 = -4/x-3

(x-3)^2=-9.-4=36

=>TH1 x-3= 6

   TH2  x-3 = -6

=> TH1: X= 9

     TH2: X=-3

=>(x-3)^2=36

=>x-3=6 hoặc x-3=-6

=>x=-3; x=9