a: ĐKXĐ: \(n\ne4\)

Để A là số nguyên thì \(3n+9⋮n-4\)

=>\(3n-12+21⋮n-4\)

=>\(21⋮n-4\)

=>\(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=>\(n\in\left\{5;3;7;1;11;-3;25;-17\right\}\)

b: ĐKXĐ: \(n\ne\dfrac{1}{2}\)

Để B là số nguyên thì \(6n+5⋮2n-1\)

=>\(6n-3+8⋮2n-1\)

=>\(8⋮2n-1\)

=>\(2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

mà 2n-1 lẻ(do n là số nguyên)

nên \(2n-1\in\left\{1;-1\right\}\)

=>\(n\in\left\{1;0\right\}\)

Chu vi tam giác đó là:

\(47+30+26=103\left(cm\right)\)

Diện tích tam giác đó là:

\(26\times13:2=169\left(cm^2\right)\)

a: Xét ΔAEC vuông tại E và ΔAHB vuông tại H có

\(\widehat{EAC}\) chung

Do đó: ΔAEC~ΔAHB

b; Xét ΔHCB vuông tại H và ΔKAC vuông tại K có

\(\widehat{HCB}=\widehat{KAC}\)(AD//BC)

Do đó: ΔHCB~ΔKAC

=>\(\dfrac{HC}{AK}=\dfrac{BC}{CA}\)

=>\(BC\cdot AK=CH\cdot CA\)

c: Xét ΔBHA vuông tại H có \(sinBAH=\dfrac{BH}{BA}\)

=>\(\dfrac{2}{BA}=sin30=\dfrac{1}{2}\)

=>BA=4(cm)

ΔAHB~ΔAEC

=>\(\dfrac{S_{AHB}}{S_{AEC}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{4}{3}\right)^2=\dfrac{16}{9}\)

a) 

\(A=\dfrac{1,11+0,19-13.2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\\ =\dfrac{1,3-26}{2,6}-\dfrac{3}{4}.\dfrac{1}{2}\\ =\dfrac{1,3\left(1-20\right)}{1,3.2}-\dfrac{3}{8}\\ =\dfrac{-19}{2}-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\\ =\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-0,5\right):\dfrac{75}{26}\\ =\left[\left(3-0,5\right)+\left(\dfrac{7}{8}-\dfrac{2}{8}\right)\right]:\dfrac{75}{26}\\ =\left(2,5+\dfrac{5}{8}\right):\dfrac{75}{26}\\ =\dfrac{25}{8}.\dfrac{26}{75}=\dfrac{13}{12}\)

b) Để \(A< x< B\) thì: \(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

\(\Rightarrow x\in\left\{-9;-8;-7;...;1\right\}\) (do \(x\in\mathbb{Z}\))

a: \(3\cdot9\cdot\left(-27\right)=3\cdot3^2\cdot\left(-3^3\right)=-3^6\)

b: \(5\cdot25\cdot\left(-125\right)^2=5\cdot5^2\cdot\left(5^3\right)^2=5^9\)

c: \(0,5\cdot\left(-0,25\right)\cdot0,0625=0,5\cdot\left(-1\right)\cdot\left(0,5\right)^2\cdot\left(0,5\right)^4\)

\(=-\left(0,5\right)^7\)

d: \(2\cdot32\cdot\left(-1024\right)=2\cdot2^5\cdot\left(-1\right)\cdot2^{10}=-2^{16}\)

e: \(49\cdot7^3\cdot\left(-7\right)^3=7^2\cdot7^3\cdot\left(-1\right)\cdot7^3=-7^8\)

f: \(\dfrac{3}{4}\cdot\dfrac{9}{16}\cdot\dfrac{27}{64}=\dfrac{3}{4}\cdot\left(\dfrac{3}{4}\right)^2\cdot\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^6\)

a, 3.9.27

= - 3.3².3³

= - 3¹⁺²⁺³ 

= - 3³⁺³

= - 3⁶

1: \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}\)

\(=-\dfrac{40}{60}+\dfrac{45}{60}+\dfrac{10}{60}-\dfrac{24}{60}\)

\(=\dfrac{5-14}{60}=-\dfrac{9}{60}=-\dfrac{3}{20}\)

2: \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{5}{6}\right)+\left(-\dfrac{1}{5}+\dfrac{7}{10}\right)\)

\(=\left(-\dfrac{8}{12}+\dfrac{9}{12}-\dfrac{10}{12}\right)+\left(-\dfrac{2}{10}+\dfrac{7}{10}\right)\)

\(=\dfrac{-9}{12}+\dfrac{5}{10}=-\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{3}{4}+\dfrac{2}{4}=-\dfrac{1}{4}\)

3: \(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\)

\(=\dfrac{3+2+1}{6}+\dfrac{14+25-4}{35}+\dfrac{1}{41}\)

\(=\dfrac{6}{6}+\dfrac{35}{35}+\dfrac{1}{41}=2+\dfrac{1}{41}=\dfrac{83}{41}\)

4: \(\dfrac{1}{100\cdot99}-\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)

\(=\dfrac{1}{100\cdot99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{97\cdot98}+\dfrac{1}{98\cdot99}\right)\)

\(=\dfrac{1}{100\cdot99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{98}{99}=\dfrac{-97}{99}-\dfrac{1}{100}=\dfrac{-9799}{9900}\)

5: \(\dfrac{\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right)\cdot\dfrac{5}{19}}{\left(\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{-3}{35}\right)\cdot\dfrac{-4}{3}}=\dfrac{\dfrac{18-16-21}{60}\cdot\dfrac{5}{19}}{\dfrac{5+10+6}{70}\cdot\dfrac{-4}{3}}\)

\(=\dfrac{\dfrac{-19}{60}\cdot\dfrac{5}{19}}{\dfrac{21}{70}\cdot\dfrac{-4}{3}}=\dfrac{-5}{60}:\dfrac{-84}{210}=\dfrac{-1}{12}\cdot\dfrac{-5}{2}=\dfrac{5}{24}\)

6: \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)

Gọi mẫu số của phân số cần tìm là x

Theo đề, ta có: \(-\dfrac{11}{13}< \dfrac{9}{x}< \dfrac{-11}{15}\)

=>\(\dfrac{11}{13}>\dfrac{-9}{x}>\dfrac{11}{15}\)

=>\(\dfrac{99}{117}>\dfrac{-99}{11x}>\dfrac{99}{135}\)

=>\(\dfrac{99}{117}>\dfrac{99}{-11x}>\dfrac{99}{135}\)

=>\(-11x\in\left\{118;119;...;134\right\}\)

=>\(x\in\left\{-\dfrac{118}{11};-\dfrac{119}{11};...;\dfrac{134}{-11}\right\}\)

mà x nguyên

nên \(x\in\left\{-11;-12\right\}\)

Vậy: Hai phân số cần tìm là \(\dfrac{9}{-11};\dfrac{9}{-12}\)

=>

\(\dfrac{a}{d}+\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{c}{d}\\ =>\dfrac{a}{b}\cdot\dfrac{c}{d}-\dfrac{c}{d}=\dfrac{a}{b}\\ =>\dfrac{c}{d}\cdot\left(\dfrac{a}{b}-1\right)=\dfrac{a}{b}\\ =>\dfrac{c}{d}\cdot\dfrac{a-b}{b}=\dfrac{a}{b}\\ =>\dfrac{c}{d}=\dfrac{a}{b}:\dfrac{a-b}{b}\\ =>\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{b}{a-b}\\ =>\dfrac{c}{d}=\dfrac{a}{a-b}\)

Vậy: ... 

a) Để A là phân số thì \(n-2\ne0\Rightarrow n\ne2\)

b) \(A=-3=>\dfrac{2n-7}{n-2}=-3\)

\(=>2n-7=-3\left(n-2\right)\\ =>2n-7=-3n+6\\ =>2n+3n=6+7\\ =>5n=13\\ =>n=\dfrac{13}{5}\left(ktm\right)\) 

c) \(A=\dfrac{2n-7}{n-2}=\dfrac{2n-4-3}{n-2}=\dfrac{2\left(n-2\right)-3}{n-2}=2-\dfrac{3}{n-2}\)

Để A nguyên thì: 3 ⋮ n - 2

=> n - 2 ∈ Ư(3) ={1; -1; 3; -3}

=> n ∈ {3; 1; 5; -1} 

d) Để A lớn nhất thì \(\dfrac{3}{n-2}\) nhỏ nhất 

=> \(\dfrac{3}{n-2}=-1\)

=>  3 = -(n - 2)

=> 3 = -n + 2

=> n = -1  

e) Để A nhỏ nhất thì \(\dfrac{3}{n-2}\) lớn nhất

=> \(\dfrac{3}{n-2}=1\)

=> 3 = n - 2

=> n = 3 + 2 

=> n = 5

f) Để A là phân số tối giản => ƯCLN(2n - 7; n - 2) = 1

=> ƯCLN(3; n - 2) = 1

=> n - 2 không chia hết cho 3

=> n - 2 ≠ 3k 

=> n ≠ 3k + 2

g) Gọi d là ước nguyên tố của 2n - 7 và n - 2 ta có: 

2n - 7 ⋮ d và n - 2 ⋮ d 

=> 2n - 7 ⋮ d và 2(n - 2) ⋮ d

=> (2n - 4)  - (2n - 7) ⋮ d

=> 3 ⋮ d 

=> d ∈ {1; -1; 3; -3}

Mà d là STN => d = 3 

Với d = 3 => 2n - 7 ⋮ 3 => 2(2n - 7) ⋮ 3 => 4n - 7 ⋮ 3 

=> 3n + n - 7 ⋮ 3 

=> n - 7 ⋮ 3 

=> n - 7 = 3k 

=> n = 3k + 7 

bạn cho mình hỏi sao câu d và câu e lại là -1 và 1 thế ạ?