Bài 3:

Gọi số cần tìm là x

Theo đề, ta có: -2x+3=-7-x

=>-2x+3=-x-7

=>-2x+x=-7-3

=>-x=-10

=>x=10

vậy: Số cần tìm là 10

Bài 1:

a: -100<x<0

mà x là số có chữ số tận cùng là 0

nên \(x\in\left\{-10;-20;-30;...;-90\right\}\)

Số số hạng là \(\dfrac{\left(-90+10\right)}{-10}+1=\dfrac{-80}{-10}+1=9\left(số\right)\)

Tổng của dãy số là \(\left(-90-10\right)\cdot\dfrac{9}{2}=-100\cdot\dfrac{9}{2}=-450\)

b: -12<=x<=20

mà x chia hết cho 5

nên \(x\in\left\{-10;-5;0;5;10;15;20\right\}\)

Tổng của dãy số là:

(-10)+(-5)+0+5+10+15+20

=(-10+10)+(-5+5)+0+15+20

=0+0+0+35

=35

c: -22<=x<14

mà x chia hết cho 9

nên \(x\in\left\{-18;-9;0;9\right\}\)

Tổng của dãy số là:

(-18)+(-9)+0+9

=(-18+0)+(-9+9)

=-18+0=-18

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 \(x\in\mathbb{Z}\))

Bài 12:

Thay x=1 vào A(x), ta được:

\(A\left(1\right)=\left(3-4\cdot1+1^2\right)^{2004}\cdot\left(3+4\cdot1+1^2\right)^{2005}=0\)

=>Tổng của tất cả các hệ số là 0

Vì \(BM=\dfrac{1}{4}BC\)

nên \(CM=\dfrac{3}{4}BC\)

AI=IM

=>I là trung điểm của AM

=>\(S_{ACM}=2\times S_{CIM}=60\left(cm^2\right)\)

Vì \(CM=\dfrac{3}{4}BC\)

nên \(\dfrac{S_{ACM}}{S_{ACB}}=\dfrac{3}{4}\)

=>\(S_{ACB}=S_{ACM}\cdot\dfrac{4}{3}=60\cdot\dfrac{4}{3}=80\left(cm^2\right)\)

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

Bài 3:

a: ĐKXĐ: \(x\ne-1\)

Để A là số nguyên thì \(4⋮x+1\)

=>\(x+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{0;-2;1;-3;3;-5\right\}\)

b: DKXĐ: x<>-1

Để B là số nguyên thì \(x+3⋮x+1\)

=>\(x+1+2⋮x+1\)

=>\(2⋮x+1\)

=>\(x+1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{0;-2;1;-3\right\}\)

c: ĐKXĐ: x<>2

Để C là số nguyên thì \(x-5⋮x-2\)

=>\(x-2-3⋮x-2\)

=>\(-3⋮x-2\)

=>\(x-2\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{3;1;5;-1\right\}\)

d: ĐKXĐ: x<>-1/2

Để D là số nguyên thì \(4x-3⋮2x+1\)

=>\(4x+2-5⋮2x+1\)

=>\(-5⋮2x+1\)

=>\(2x+1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{0;-1;2;-3\right\}\)

Bài 4:

a: ĐKXĐ: \(x\ne0\)

Để \(\dfrac{3}{x}>0\) thì x>0

b: ĐKXĐ: \(x\ne0\)

Để \(\dfrac{4}{3x}>0\) thì 3x>0

=>x>0

c: ĐKXĐ: \(x\ne-1\)

Để \(\dfrac{2}{x+1}>0\) thì x+1>0

=>x>-1

d: ĐKXĐ: \(x\ne2\)

Để \(-\dfrac{1}{x-2}\)>0 thì x-2<0

=>x<2

e: ĐKXĐ: \(x\ne-4\)

Để \(\dfrac{x}{x+4}>0\) thì \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>0\\x< -4\end{matrix}\right.\)

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

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

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