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

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

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

Gọi số cần tìm có dạng là \(\overline{ab}\)

Nếu viết thêm chữ số 0 vào giữa hai chữ số thì ta được số mới gấp 6 lần số cũ nên \(\overline{a0b}=6\cdot\overline{ab}\)

=>\(100a+b=6\cdot\left(10a+b\right)\)

=>100a+b=60a+6b

=>40a=5b

=>8a=b

=>b=8; a=1

Vậy: Số cần tìm là 18

Là sai chứ sao

a: Để hệ có nghiệm duy nhất thì \(\dfrac{3}{2}\ne\dfrac{a}{1}\)

=>\(a\ne1,5\)

b: Để hệ vô nghiệm thì \(\dfrac{3}{2}=\dfrac{a}{1}\ne\dfrac{5}{b}\)

=>\(\left\{{}\begin{matrix}a=\dfrac{3}{2}\\b\ne\dfrac{10}{3}\end{matrix}\right.\)

c: Để hệ có vô số nghiệm thì \(\dfrac{3}{2}=\dfrac{a}{1}=\dfrac{5}{b}\)

=>\(\left\{{}\begin{matrix}a=1\cdot\dfrac{3}{2}=\dfrac{3}{2}\\b=5\cdot\dfrac{2}{3}=\dfrac{10}{3}\end{matrix}\right.\)

Bài 7:

a: \(\dfrac{11}{13}=\dfrac{110}{130};\dfrac{12}{13}=\dfrac{120}{130}\)

=>4 phân số nằm giữa 11/13 và 12/13 là \(\dfrac{111}{130};\dfrac{112}{130};\dfrac{113}{130};\dfrac{116}{130}\)

b: \(\dfrac{15}{17}=\dfrac{150}{170};\dfrac{15}{16}=\dfrac{150}{160}\)

=>5 phân số nằm giữa 15/17 và 15/16 là \(\dfrac{150}{169};\dfrac{150}{168};\dfrac{150}{167};\dfrac{150}{165};\dfrac{150}{153}\)

Bài 11:

\(a)x\times12,8=6,4\\ x=6,4:12,8\\ x=\dfrac{1}{2}\\ b)17,3:x=69,2\\ x=17,3:69,2\\ x=\dfrac{1}{4}\\ c)16,48\times x=4,12\\ x=4,12:16,48\\ x=\dfrac{1}{4}\\ d)x:12,8=1,6\\ x=12,8\times1,6\\ x=\dfrac{512}{25}\)

Bài 11:

a: \(x\times12,8=6,4\)

=>\(x=\dfrac{6.4}{12.8}=\dfrac{1}{2}=0,5\)

b: \(17,3:x=69,2\)

=>\(x=\dfrac{17.3}{69.2}=0,25\)

c: \(16,48\times x=4,12\)

=>\(x=\dfrac{4.12}{16.48}=\dfrac{1}{4}=0,25\)

d: \(x:12,8=1,6\)

=>\(x=12,8\times1,6=20,48\)

Bài 9:

a: \(4,86\times0,25\times40=4,86\times10=48,6\)

b: \(0,125\times6,94\times80=6,94\times\left(80\times0,125\right)\)

=6,94x10

=69,4

c: \(96,28\times3,527+3,527\times3,72\)

=3,527x(96,28+3,72)

=3,527x100=352,7

d: \(72,9\times99+72+0,9\)

=72,9x99+72,9

=72,9x(99+1)

=72,9x100=7290

e: \(0,8\times96+1,6\times2\)

\(=0,8\times96+0,8\times4=0,8\times\left(96+4\right)\)

=0,8x100=80

\(D=x^2+y^2-x+6y+10\\ =\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(y^2+2\cdot y\cdot3+3^2\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.=>D=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

Dấu "=" xảy ra \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

______________________________

\(F=2xy-2x^2-y^2+10x-27\\ =-\left(x^2-2xy+y^2\right)-\left(x^2-10x+25\right)-2\\ =-\left(x-y\right)^2-\left(x-5\right)^2-2\)

Ta có: \(\left\{{}\begin{matrix}\left(x-y\right)^2\le0\forall x,y\\-\left(x-5\right)^2\le0\forall x\end{matrix}\right.=>F=-\left(x-y\right)^2-\left(x-5\right)^2-2\le-2\forall x,y\)

Dấu "=" xảy ra: \(\left\{{}\begin{matrix}x-y=0\\x-5=0\end{matrix}\right.\Leftrightarrow x=y=5\)

\(A=-x^2+x-1\)

\(=-\left(x^2-x+1\right)\)

\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

=>\(x=\dfrac{1}{2}\)

\(B=6x-x^2-10\)

\(=-\left(x^2-6x+10\right)\)

\(=-\left(x^2-6x+9+1\right)\)

\(=-\left(x-3\right)^2-1< =-1\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

\(C=-x^2+5x+3\)

\(=-\left(x^2-5x-3\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}< =\dfrac{37}{4}\forall x\)

Dấu '=' xảy ra khi x-5/2=0

=>x=5/2

\(D=x^2-x+y^2+6y+10\)

\(=x^2-x+\dfrac{1}{4}+y^2+6y+9+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

\(F=2xy-2x^2-y^2+10x-27\)

\(=-\left(2x^2+y^2-2xy-10x+27\right)\)

\(=-\left(x^2-2xy+y^2+x^2-10x+25+2\right)\)

\(=-\left(x-y\right)^2-\left(x-5\right)^2-2< =-2\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=x=5\end{matrix}\right.\)