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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\frac{a^{1994}}{b^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\)(1)
\(\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}=\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}\)(2)
từ (1) và (2) => \(\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\left(đpcm\right)\)
\(\)
\(\frac{a}{b}=\frac{c}{d}\)=\(\frac{a+c}{b+d}\)
=> \(\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}\)\(=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\)
=> \(\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\)
=> dpcm
$\left ( a+b\sqrt{2} \right )^{1994}+\left ( c+d\sqrt{2} \right )^{1994}= 5+4\sqrt{2}$ - Đại số - Diễn đàn Toán học
Đặt \(x-\frac{a+b}{2}=X\)
\(\Rightarrow y=\left(X-\frac{a-b}{2}\right)^{1994}+\left(X+\frac{a-b}{2}\right)^{1994}\)
\(y\left(-X\right)=\left(-X-\frac{a-b}{2}\right)^{1994}+\left(-X+\frac{a-b}{2}\right)^{1994}\)
\(=\left(X+\frac{a-b}{2}\right)^{1994}+\left(X-\frac{a-b}{2}\right)^{1994}=y\left(X\right)\)
\(\Rightarrow y\left(X\right)\) là hàm chẵn \(\Rightarrow\) đồ thị hàm số đối xứng qua trục \(X=0\) hay đồ thị hàm \(y\left(x\right)\) đối xứng qua trục \(x-\frac{a+b}{2}=0\Leftrightarrow x=\frac{a+b}{2}\)
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}=\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}=\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}\)
=> Đpcm
Câu 2 tớ đăng phía dưới rồi đó.
Câu 3 đang định đăng lên thì cậu đăng là sao hả?