C/m rằng nếu : a+b+v+d = 0 thì \(a^3+b^3+c^3+d^3=3\left(a.c-b.d\right)\) \(.\left(b+d\right)\)
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TN
cho \(b^2=a.c-a^2=b.d\)
c/m:\(\dfrac{a^3+b^3-c^3}{b^3+c^2-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^2\)
2
23 tháng 10 2018
\(b^2=a.c\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=b.d\)
\(\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
\(=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}\left(đpcm\right)\)
NV
0
12 tháng 2 2018
Ta có: \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{a^3+b^3+c^3+2ab+2ac+2bc}{b^3+c^3+d^3+2bc+2bd+2cd}\)
NT
0
13 tháng 7 2016
Ta có : \(a+b+c+d=0\Leftrightarrow a+d=-\left(b+c\right)\)
\(\Leftrightarrow\left(a+d\right)^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[c^3+b^3+3bc\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ad\left(b+c\right)-3bc\left(b+c\right)\) (vì a + d = - b - c )
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
21 tháng 12 2017
1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
21 tháng 12 2017
3. \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
\(\text{Mà }a-b=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
a +b+c+d=0
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