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a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
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Bài 1 :
\(-8=\frac{-8}{1}=\frac{-16}{2}=\frac{-24}{3}=\frac{-32}{4}=\frac{-40}{5}\)
\(-2=\frac{-2}{1}=\frac{-4}{2}=\frac{-6}{3}=\frac{-8}{4}=\frac{-10}{5}\)
\(3=\frac{3}{1}=\frac{6}{2}=\frac{9}{3}=\frac{12}{4}=\frac{15}{5}\)
Bài 2 :
a) Để A là phân số thì :
\(n-6\ne0\Rightarrow n\ne6\)
b)\(A=\frac{4}{0-6}=\frac{4}{-6}\)
\(A=\frac{4}{7-6}=4\)
\(A=\frac{4}{-12-6}=\frac{-2}{9}\)
Bài 3 : [ Tương tự bài 2 ]
Bài 4 : [ Suy nghĩ thì ra ]
[ Hoq chắc - có gì sai thông cảm ]
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Đề sai rồi! Sửa đề: Cho \(S_1=\dfrac{b}{a}x+\dfrac{c}{a}z...\)
Giải:
Ta có:
\(S_1+S_2+S_3=\left(\dfrac{b}{a}x+\dfrac{c}{a}z\right)+\left(\dfrac{a}{b}x+\dfrac{c}{b}y\right)\)\(+\left(\dfrac{a}{c}z+\dfrac{b}{c}y\right)\)
\(=\left(\dfrac{b}{a}x+\dfrac{a}{b}x\right)+\left(\dfrac{c}{b}y+\dfrac{b}{c}y\right)+\left(\dfrac{c}{a}z+\dfrac{a}{c}z\right)\)
\(=\left(\dfrac{b}{a}+\dfrac{a}{b}\right)x+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)y+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)z\)
Dễ thấy: \(\left\{{}\begin{matrix}\dfrac{b}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{b}+\dfrac{b}{c}\ge2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\end{matrix}\right.\)
\(\Rightarrow S_1+S_2+S_3\ge2x+2y+2z\)
\(=2\left(x+y+z\right)=2.1008=2016\)
Vậy \(S_1+S_2+S_3\ge2016\) (Đpcm)
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a/ ta có:\(\frac{x}{-3}=\frac{4}{y}\Rightarrow xy=-12\)
mà x;y nguyên nên ta có
th1: \(x=\pm1;y=\pm12\) th2\(x=\pm12;y=\pm1\)...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{8}=\frac{-15}{30}\)
\(\Rightarrow x=\frac{-15.8}{30}\)
\(\Rightarrow x=-4\)
\(\frac{y}{6}=\frac{15}{-45}\)
\(\Rightarrow y=\frac{15.6}{-45}\)
\(\Rightarrow y=-2\)
A) \(\frac{x}{8}=\frac{-15}{30}\)
\(\Leftrightarrow\)\(x=\frac{\left(-15\right).8}{30}=-4\)
Vậy...
B) \(\frac{y}{6}=\frac{15}{-45}\)
\(\Leftrightarrow\)\(y=\frac{15.6}{-45}=-2\)
Vậy....
ta có:a-x/b-y=a/b
=>(a-x).b=(b-y).a
=>ab-xb=ab-ay
do đó xb=ay(rút gọn ab)
=>x/y=a/b(đpcm)