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3.
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Leftrightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\) và \(a+2b-3c=-20\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)
+) \(\dfrac{a}{2}=5\Rightarrow a=5.2=10\)
+) \(\dfrac{2b}{6}=5\Rightarrow2b=5.6=30\Rightarrow b=30:2=15\)
+) \(\dfrac{3c}{12}=5\Rightarrow3c=5.12=60\Rightarrow c=60:3=20\)
Vậy ...
3.
ta có:\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=>\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\) và a+2b-3c=-20
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\)=\(\dfrac{a+2b-3c}{2+6-12}\)\(\dfrac{-20}{-4}\)=5
vì\(\dfrac{a}{2}\)=5=>a=2.5=10
\(\dfrac{2b}{6}\)=5=>2b=5.6=30=>b=30:2=15
\(\dfrac{3c}{12}\)=5=>3c=5.12=60=>c=60:3=20
vậy a=10,b=15,c=20
chúc bạn hok tốt
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Bài 2:
a: \(3B=3+3^2+3^3+...+3^{90}\)
\(\Leftrightarrow2B=3^{90}-1\)
hay \(B=\dfrac{3^{90}-1}{2}\)
b: \(B=\left(1+3+3^2+3^3+3^4+3^5\right)+3^6\left(1+3+3^2+3^3+3^4+3^5\right)+...+3^{84}\left(1+3+3^2+3^3+3^4+3^5\right)\)
\(=384\cdot\left(1+3^6+...+3^{84}\right)⋮52\)
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1
a) Vì \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)
\(\Rightarrow ad< bc\)
2
b) Ta có : \(\dfrac{-1}{3}=\dfrac{-16}{48};\dfrac{-1}{4}=\dfrac{-12}{48}\)
Ta có dãy sau : \(\dfrac{-16}{48};\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48};\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa \(\dfrac{-1}{3}\) và \(\dfrac{-1}{4}\) là :\(\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}\)
1a ) Ta có : \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\)
\(\Leftrightarrow\) \(\dfrac{ad}{bd}\) < \(\dfrac{bc}{bd}\) \(\Rightarrow\) ad < bc
1b ) Như trên
2b) \(\dfrac{-1}{3}\) = \(\dfrac{-16}{48}\) ; \(\dfrac{-1}{4}\) = \(\dfrac{-12}{48}\)
\(\dfrac{-16}{48}\) < \(\dfrac{-15}{48}\) <\(\dfrac{-14}{48}\) < \(\dfrac{-13}{48}\) < \(\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa là.................
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2: \(A=9^n\cdot81-9^n+3^n\cdot9+3^n\)
\(=9^n\cdot80+3^n\cdot10\)
\(=10\left(9^n\cdot8+3^n\right)⋮10\)
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1
\(\frac{x-3}{4}=\frac{y+5}{3}=\frac{z-4}{5}=\frac{2x-6}{8}=\frac{3y+15}{9}=\frac{4z-16}{20}\)
\(=\frac{2x+3y-4z-6+15+16}{-3}=-\frac{100}{3}\)
Làm nốt
2
\(\left|x-2\right|\ge0\) dấu "=" xảy ra tại x=2
\(\left(x-y\right)^2\ge0\) dấu "=" xảy ra tại x=y
\(3\sqrt{z^2+9}\ge3\sqrt{9}=9\) dấu "=" xảy ra tại z=0
\(\Rightarrow C\ge0+0+9+16=25\) dấu "=" xảy ra tại x=y=2;z=0
5
Chứng minh \(1< M< 2\) là OK