4/a = 9/a tìm a
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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
`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`
\(\left(A+9\right)+\left(A-8\right)+\left(A+7\right)+\left(A-6\right)+\left(A+5\right)+\left(A-4\right)=63.3\)
\(\Leftrightarrow\left(A+A+A+A+A+A\right)+\left(9-8+7-6+5-4\right)=189\)
\(\Leftrightarrow6A+3=189\)
\(\Leftrightarrow6A=186\)
\(\Leftrightarrow A=31\)
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 \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
1. Tìm x, y biết: 6x/3y=4/9 và 3x=9-y
2. Tìm a,b,c biết a/2=b/3=c/4 và a+2b+20-3c=0
GIÚP MÌNH NHA !!!!
4/4 = 9/9