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\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
=> a2 = 16
=> a = 4 hoặc a = -4
Thay vào \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) tìm nốt a, b, c
hjhj, thật ra bài này mik làm đc. mik gửi cho vui thôi
dù gì thì
1. \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}a=5\times2=10\\b=5\times3=15\\c=5\times4=20\end{matrix}\right.\)
tham khảo!!
https://lazi.vn/edu/exercise/tim-cac-so-a-b-c-biet-rang-a-2-b-3-c-4-va-a-2-b-2-2c-2-108
a)a:b:c=2:4:5 =>\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{2a}{4}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{2a-b+c}{4-4+5}=\dfrac{7}{5}\)
=>a=\(2\cdot\dfrac{7}{5}=\dfrac{14}{5}\)
\(b=4\cdot\dfrac{7}{5}=\dfrac{28}{5}\)
\(c=5\cdot\dfrac{7}{5}=7\)
Vậy...
b)\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
=>a2=16 b2=36 c2=64
=>a=4 b=6 c=8 hoặc a=-4 b=-6 c=-8
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{15};\dfrac{b}{5}=\dfrac{c}{4}\Rightarrow\dfrac{b}{15}=\dfrac{c}{12}.\)
Do đó : \(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}=\dfrac{a-b+c}{10-15+12}=\dfrac{-49}{7}=-7.\)
\(\Rightarrow a=-70;b=-105;c=-84.\)
Theo đề bài: \(\dfrac{a}{2}=\dfrac{b}{3}\); \(\dfrac{b}{5}=\dfrac{c}{4}\)
\(\Rightarrow\) \(\dfrac{a}{10}=\dfrac{b}{15}\); \(\dfrac{b}{15}=\dfrac{c}{12}\)
\(\Rightarrow\) \(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}=\dfrac{a-b+c}{10-15+12}=\dfrac{-49}{7}=-7\)
\(\Rightarrow\dfrac{a}{10}=-7\Rightarrow a=-70\)
và \(\dfrac{b}{15}=-7\Rightarrow b=-105\)
và \(\dfrac{c}{12}=-7\Rightarrow c=-84\)
Vậy \(a=-70\); \(b=-105\); \(c=-84\)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)
Ta có :
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\left\{\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=\left(-16\right)\)
\(\Rightarrow4k^2+27k^2-32k^2=-16\)
\(\Rightarrow\left(-1\right)k^2=-16\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow a=8;b=12;c=16\)
+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4.\)
Mình nghĩ bạn t hiếu cái gì đó thì phải