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\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\\ \dfrac{2k}{2}=\dfrac{3k}{3}=\dfrac{4k}{4}\\ \Rightarrow\dfrac{\left(2k\right)^2}{2^2}=\dfrac{\left(3k\right)^2}{3^2}=\dfrac{2\left(4k\right)^2}{2\cdot4^2}\\ \Leftrightarrow\dfrac{4k^2}{4}=\dfrac{9k^2}{9}=\dfrac{32k^2}{32}=\dfrac{4k^2-9k^2+32k^2}{4-9+32}=\dfrac{108}{27}=4\\ \dfrac{4k^2-9k^2+32k^2}{4-9+32}=4\\ \Rightarrow\dfrac{\left(4-9+32\right)k^2}{4-9+32}=4\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ k=2\Rightarrow\left\{{}\begin{matrix}a=2k=2\cdot2=4\\b=3k=3\cdot2=6\\c=4k=4\cdot2=8\end{matrix}\right.\\ k=-2\Rightarrow\left\{{}\begin{matrix}a=2k=2\cdot\left(-2\right)=-4\\b=3k=3\cdot\left(-2\right)=-6\\c=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)
Vậy ...
Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau có :
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{a}{2}=4\\\dfrac{b}{3}=4\\\dfrac{c}{4}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=8\\b=12\\c=16\end{matrix}\right.\)
e: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)
=>x+5=-34; y-2=-51
=>x=-39; y=-49
g: Áp dụng tính chất của DTSBN, ta được
\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)
=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13
=>a=-493/13; b=-1051/13; c=-1453/13
Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$
Suy ra:
$x+5=2(-17)=-34\Rightarrow x=-39$
$y-2=3(-17)=-51\Rightarrow y=-49$
f. Đề thiếu. Bạn xem lại
h. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$
$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$
$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$
$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$
Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$
$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$
$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$
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
1. \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}a=5\times2=10\\b=5\times3=15\\c=5\times4=20\end{matrix}\right.\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{c}{a}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
Mà \(a=2005\Rightarrow b=c=2005\)
Vậy \(b=c=2005\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
=> 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
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4.\)
Câu 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2}{c^2}=\dfrac{b^2k^2}{d^2k^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2a^2+3b^2}{2c^2+3d^2}=\dfrac{2b^2k^2+3b^2}{2d^2k^2+3d^2}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{2a^2+3b^2}{2c^2+3d^2}\)
b: \(\dfrac{2a-3c}{c}=\dfrac{2bk-3dk}{dk}=\dfrac{2b-3d}{d}\)
4. \(1^2+2^2+3^2+...+10^2+11^2=506\)
Ta có: \(2^2+4^2+6^2+...+20^2+22^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2+2^2.11^2\)
\(=2^2\left(1^2+2^2+3^2+...+10^2+11^2\right)\)
\(=2^2.506=2024\)
Vậy....
1.
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow a^2=16\)
\(\Rightarrow b^2=36\)
\(\Rightarrow c^2=64\)
\(\Rightarrow a=\pm4\) , \(b=\pm6\) , \(c=\pm8\)
Lời giải:
a.
Đặt $\frac{a}{5}=\frac{b}{4}=k\Rightarrow a=5k, b=4k$
Khi đó:
$a^2-b^2=1$
$\Rightarrow (5k)^2-(4k)^2=1$
$\Rightarrow 9k^2=1\Rightarrow k^2=\frac{1}{9}\Rightarrow k=\frac{1}{3}$ hoặc $k=\frac{-1}{3}$
Nếu $k=\frac{1}{3}$ thì:
$a=5k=\frac{5}{3}; b=4k=\frac{4}{3}$
Nếu $k=\frac{-1}{3}$ thì:
$a=5k=\frac{-5}{3}; b=4k=\frac{-4}{3}$
b.
Đặt $\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k$
$\Rightarrow a=2k; b=3k; c=4k$
Khi đó:
$a^2-b^2+2c^2=108$
$\Rightarrow (2k)^2-(3k)^2+2(4k)^2=108$
$\Rightarrow 27k^2=108$
$\Rightarrow k^2=4\Rightarrow k=\pm 2$
Nếu $k=2$ thì:
$a=2k=4; b=3k=6; c=4k=8$
Nếu $k=-2$ thì:
$a=2k=-4; b=3k=-6; c=4k=-8$