Lấy máy tính tính đi, câu này dễ mà còn hỏi cho bằng được nữa =="

\(\dfrac{5.6}{9.35}=\dfrac{5.3.2}{3.3.5.7}=\dfrac{2}{3.7}=\dfrac{2}{21}\)

\(\dfrac{7.2+8}{2.14.5}=\dfrac{7.2+8.2}{2.7.2.5}=\dfrac{\left(7+8\right).2}{2.7.2.5}=\dfrac{15.2}{2.7.2.5}=\dfrac{3.5.2}{2.7.2.5}=\dfrac{3}{14}\)

1,\(\dfrac{5.6}{9.35}\) = \(\dfrac{5.2.3}{3.3.5.7}\) = \(\dfrac{2}{3.7}\)

2, \(\dfrac{7.2+8}{2.14.5}\) = \(\dfrac{7.2+2.4}{2.14.5}\) = \(\dfrac{2\left(7+4\right)}{2\left(14.5\right)}\) = \(\dfrac{2.11}{2.70}\)= \(\dfrac{11}{70}\)

\(\dfrac{7.2+4.2}{2.14.5}=\dfrac{2\left(7+4\right)}{2.14.5}\)=\(\dfrac{11}{70}\)

\(a,\dfrac{7}{35},\dfrac{18}{54},\dfrac{-15}{125},\dfrac{-4}{25}\)

Các thừa số đã tối giản : \(\dfrac{-4}{25}\)

\(\dfrac{7}{35}=\dfrac{7:7}{35:7}=\dfrac{1}{5}\) , \(\dfrac{18}{54}=\dfrac{18:18}{54:18}=\dfrac{1}{3}\)

\(\dfrac{-15}{125}=\dfrac{-15:5}{125:5}=\dfrac{-3}{25}\)

\(b,\dfrac{27}{45},\dfrac{21}{28},\dfrac{8}{14},\dfrac{18}{-60},\dfrac{-270}{360}\)

Các thừa số đã tối giản là : ko có

\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\) , \(\dfrac{21}{28}=\dfrac{21:7}{28:7}=\dfrac{3}{4}\)

\(\dfrac{8}{14}\)\(=\dfrac{8:2}{14:2}=\dfrac{4}{7}\) , \(\dfrac{18}{-60}=\dfrac{18:6}{-60:6}=\dfrac{3}{-10}=\dfrac{-3}{10}\)

\(\dfrac{-270}{360}=\dfrac{-270:90}{360:90}=\dfrac{-3}{4}\)

\(c,\dfrac{3.4+3.7}{6.5+9}\) = \(\dfrac{3.\left(4+7\right)}{30+9}\) = \(\dfrac{3.11}{39}\) = \(\dfrac{3.11}{3.13}=\dfrac{11}{13}\)

\(\dfrac{-63}{81},\dfrac{9.6}{9.35},\dfrac{7.2+8}{2.14.5}\)

Các p/s đã tối giản : ko có

\(\dfrac{-63}{81}=\dfrac{-63:9}{81:9}=\dfrac{-7}{9}\) , \(\dfrac{9.6}{9.35}=\dfrac{6}{35}\)

\(\dfrac{7.2+8}{2.14.5}=\dfrac{14+8}{28.5}=\dfrac{22}{140}=\dfrac{11}{70}\)

Bài 1:

Bài 2:

Các bạn giúp mình câu này nha! 

1: B là số nguyên

=>n-3 thuộc {1;-1;5;-5}

=>n thuộc {4;2;8;-2}

3:

a: -72/90=-4/5
b: 25*11/22*35

\(=\dfrac{25}{35}\cdot\dfrac{11}{22}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)

c: \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)

a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)

b: P>=-1/2

=>P+1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)

=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)

=>căn x-9>=0

=>x>=81

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>-6/căn x+3>=-2

Dấu = xảy ra khi x=0

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)

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

e: ĐKXĐ: x<>-y

\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)