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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. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
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:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\)\(\Rightarrow\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{bk^{2014}+b^{2014}}{dk^{2014}+d^{2014}}=\dfrac{b\left(k^{2014}+b^{2013}\right)}{d\left(k^{2014}+d^{2013}\right)}\)
2 cái này thấy nó ko giống nhau lắm:v
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có:+) \(\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\) (1)
+) \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}.k^{2014}+b^{2014}}{d^{2014}.k^{2014}+d^{2014}}\)
\(=\dfrac{b^{2014}.\left(k^{2014}+1\right)}{d^{2014}.\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) => đpcm