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12 tháng 2 2023

4/4 = 9/9 

 

 

DT
15 tháng 6 2023

a) \(\dfrac{7}{4}< \dfrac{a}{8}< 3\\ =>\dfrac{7}{4}.8< a< 3.8\\ =>14< a< 24\\ =>a\in\left\{15;16;17;...;23\right\}\)

b) \(\dfrac{2}{3}< \dfrac{a-1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}.6< a-1< \dfrac{8}{9}.6\\ =>4< a-1< \dfrac{16}{3}\\ =>4+1< a< \dfrac{16}{3}+1\\ =>5< a< \dfrac{19}{3}\\ =>a=6\)

b) \(\dfrac{2}{3}< a-\dfrac{1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}+\dfrac{1}{6}< a< \dfrac{8}{9}+\dfrac{1}{6}\\ =>\dfrac{5}{6}< a< \dfrac{19}{18}\\ =>a=1\)

c) \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\\ =>\dfrac{24}{18}< \dfrac{24}{6a}< \dfrac{24}{9}\\ =>9< 6a< 18\\ =>\dfrac{9}{6}< a< \dfrac{18}{6}\\ =>1,5< a< 3\\ =>a=2\)

a: =>4/9*a/b=5/27+3/27=8/27

=>a/bb=8/27:4/9=8/27*9/4=72/108=2/3

b: =>9/13*a/b=40/91+1/7=53/91

=>a/b=53/63

11 tháng 3 2023

`a)4/9xxa/b-1/9=5/27`

`=>4/9xxa/b=5/27+1/9`

`=>4/9xxa/b=5/27+3/27`

`=>4/9xxa/b=8/27`

`=>a/b=8/27:4/9`

`=>a/b=8/27xx9/4`

`=>a/b=2/3`

Vậy `a=2` và `b=3`

__

`b)9/13xxa/b+1/7=40/91`

`=>9/13xxa/b=40/91-1/7`

`=>9/13xxa/b=40/91-13/91`

`=>9/13xxa/b=27/91`

`=>a/b=27/91:9/13`

`=>a/b=27/91xx13/9`

`=>a/b=39/91`

Vậy `a=39` và `b=91`

13 tháng 5 2018

\(\left(A+9\right)+\left(A-8\right)+\left(A+7\right)+\left(A-6\right)+\left(A+5\right)+\left(A-4\right)=63.3\)

\(\Leftrightarrow\left(A+A+A+A+A+A\right)+\left(9-8+7-6+5-4\right)=189\)

\(\Leftrightarrow6A+3=189\)

\(\Leftrightarrow6A=186\)

\(\Leftrightarrow A=31\)

Vậy A = 31

_Chúc bạn học tốt_

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

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

Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)

\(\Leftrightarrow x+2\sqrt{x}-3=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)