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Giả sử C(x)=0
--> x2 -5x + 6=0
--> x2-x+6x-6=0
--> x(x-1)+ 6(x-1)=0
-->(x+6)(x-1)=0
--> x=-6 và x=1 là các nghiệm của C(x)
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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\(\frac{-1}{5}< \frac{-1}{4}< \frac{-1}{3}< \frac{-1}{2}< -1< 0< \frac{1}{5}\)
\(\frac{-1}{2}=\frac{\left(-1\right).12}{2.12}=\frac{-12}{24}\)
\(\frac{-1}{3}=\frac{\left(-1\right).8}{3.8}=\frac{-8}{24}\)
\(\frac{-8}{24}< x< \frac{-12}{24}\)
\(\Rightarrow x=\left\{\frac{-9}{24};\frac{-10}{24};\frac{-11}{24}\right\}\)
=\(1+\widehat{ANK}\)+CO2+Từ ghép+\(\cos+\varnothing+ℕ^∗+...\approx Clgt!\) +the present simple
a)\(-\dfrac{2}{15}-x=-\dfrac{3}{10}\)
\(x=\left(-\dfrac{2}{15}\right)-\left(\dfrac{-3}{10}\right)\)
\(x=\left(-\dfrac{2}{15}\right)+\dfrac{3}{10}\)
\(x=\dfrac{1}{6}\)
a, \(\dfrac{-2}{15}-x=\dfrac{-3}{10}\)=> x=\(\dfrac{-2}{15}-\dfrac{-3}{10}=>x=\dfrac{1}{6}\)
b,\(\left(\dfrac{2x}{5}-1\right):\left(-5\right)=\dfrac{1}{4}\)
=> \(\dfrac{2x}{5}-1=\dfrac{1}{4}.\left(-5\right)\)=> \(\dfrac{2x}{5}-1=\dfrac{-5}{4}=>\dfrac{2x}{5}=\dfrac{-1}{4}=>\dfrac{8x}{20}=\dfrac{-5}{20}\)=>x=\(\dfrac{-5}{8}\)
c, \(2\dfrac{1}{4}x-9\dfrac{1}{4}=20=>\dfrac{9}{4}x-\dfrac{37}{4}=20\)=>\(\dfrac{9}{4}x=20+\dfrac{37}{4}=>\dfrac{9}{4}x=\dfrac{117}{4}=>x=13\)
d, \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)=> \(\dfrac{1}{7}:x=\dfrac{3}{7}-\dfrac{3}{14}=>\dfrac{1}{7}:x=\dfrac{3}{14}=>x=\dfrac{2}{3}\)
e, \(\dfrac{-15}{12}x+\dfrac{3}{7}=\dfrac{6}{5}x-\dfrac{1}{2}\)
=>\(\dfrac{-15}{12}x-\dfrac{6}{5}x=\dfrac{-1}{2}-\dfrac{3}{7}\)=>\(x.\left(\dfrac{-15}{12}-\dfrac{6}{5}\right)=\dfrac{-13}{14}\)
=>\(x.\dfrac{-49}{20}=\dfrac{-13}{14}=>x=\dfrac{130}{343}\)
Còn bài g thì mk k bt, hình như sai đề....