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a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
=>(a+5)(b-6)=(a-5)(b+6)
=>ab-6a+5b-30=ab+6a-5b-30
=>-6a+5b=6a-5b
=>-12a=-10b
=>6a=5b
=>\(\dfrac{a}{b}=\dfrac{5}{6}\)
b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
Ta có:
- \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)
=>ad+ab<bc+ab
=>a(b+d)>b(a+c)
=>\(\frac{a}{b}< \frac{a+c}{b+d}\) (1)
- \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)
=>ad+cd<bc+cd
=>d(a+c)<c(b+d)
=>\(\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) và (2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)(đpcm)
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\(\frac{-1}{3}=\frac{-8}{24}>\frac{-9}{24}>\frac{-10}{24}>\frac{-11}{24}>\frac{-12}{24}=\frac{-1}{2}\)
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\(\frac{-1}{5}< \frac{-1}{4}< \frac{-1}{3}< \frac{-1}{2}< -1< 0< \frac{1}{5}\)
\(\frac{-1}{2}=\frac{\left(-1\right).12}{2.12}=\frac{-12}{24}\)
\(\frac{-1}{3}=\frac{\left(-1\right).8}{3.8}=\frac{-8}{24}\)
\(\frac{-8}{24}< x< \frac{-12}{24}\)
\(\Rightarrow x=\left\{\frac{-9}{24};\frac{-10}{24};\frac{-11}{24}\right\}\)