rút gọn các biểu thức sau
a)9^9 + 27^7 / 9^6 +243^3
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Đặt \(k=\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{-3}\)
\(\Rightarrow x=2k;y=2k;z=-3k\)
Ta thay \(x=2k;y=2k;z=-3k\) vào \(xyz=240\)
\(\Rightarrow\left(2k\right)\left(2k\right)\left(-3k\right)=240\)
\(\Rightarrow\left[2.2.\left(-3\right)\right]\left(k.k.k\right)=240\)
\(\Rightarrow\left(-12k^3\right)=240\)
\(\Rightarrow k=-\sqrt[3]{20}\)
Vậy \(\left\{{}\begin{matrix}x=y=-2\sqrt[3]{20}\\z=3\sqrt[3]{20}\end{matrix}\right.\)
\(x-y-z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\)
Lại có :\(C=\left(\dfrac{1-z}{x}\right)\left(\dfrac{1-x}{y}\right)\left(\dfrac{1+y}{z}\right)\)
\(C=\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)
\(C=\dfrac{y}{x}\cdot\) \(\left(-\dfrac{z}{y}\right)\) \(\dfrac{x}{z}\)
\(C=\dfrac{-xyz}{xyz}=-1\)
\(\dfrac{x}{3}=\dfrac{y}{5}\) \(\Leftrightarrow x=\dfrac{3}{5}y\)
Thay vào B có
\(B=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(\dfrac{3}{5}y\right)^2+3y^2}{10\cdot\left(\dfrac{3}{5}y\right)^2-3y^2}=\dfrac{5\cdot\dfrac{9}{25}y^2+3y^2}{10\cdot\dfrac{9}{25}y^2-3y^2}\)
\(=\dfrac{\dfrac{9}{5}y^2+3y^2}{\dfrac{18}{5}y^2-3y^2}=\dfrac{y^2\left(\dfrac{9}{5}+3\right)}{y^2\left(\dfrac{18}{5}-3\right)}=\dfrac{\dfrac{9}{5}+3}{\dfrac{18}{5}-3}=\dfrac{\dfrac{24}{5}}{\dfrac{3}{5}}\)
\(=\dfrac{24}{5}\cdot\dfrac{5}{3}=8\)
a, \(\dfrac{7}{20}\) = \(\dfrac{8}{20}\) + (\(\dfrac{-1}{20}\))
b, \(\dfrac{7}{20}\) = \(\dfrac{1}{4}\) + \(\dfrac{1}{10}\)
a, \(\dfrac{13}{18}\) = \(\dfrac{5}{18}\) + \(\dfrac{4}{9}\)
b, \(\dfrac{13}{18}\) = \(\dfrac{14}{18}\) + ( \(\dfrac{-1}{18}\))
c, \(\dfrac{13}{18}\) = \(\dfrac{7}{18}\) + \(\dfrac{1}{3}\)
Góc xMz = tMz-xMt => xMz= 90-70 = 20 ( độ)
Góc tMy = xMy-xMt => 90-70 = 20 ( độ)
Góc zMy = xMy - zMx => 90-20 = 70 ( độ)
\(\dfrac{9^9+27^7}{9^6+243^3}=\dfrac{\left(3^2\right)^9+\left(3^3\right)^7}{\left(3^2\right)^6+\left(3^5\right)^3}=\dfrac{3^{18}+3^{21}}{3^{12}+3^{15}}=\dfrac{3^{18}\left(1+3^3\right)}{3^{12}\left(1+3^3\right)}\) \(=\dfrac{3^{18}}{3^{12}}=3^6\)