Đáp án:
1) a) $\dfrac {1}{101}$
b) x=2;y=3;z=5
x=-2;y=-3;z=-5
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
a)\,{4^{25}}.\dfrac{1}{{2.3}}.\dfrac{3}{{2.5}}.\dfrac{5}{{2.7}}.\dfrac{7}{{2.9}}...\dfrac{{99}}{{2.101}}\\
= {4^{25}}.\dfrac{{1.3.5.7.....97.99}}{{{2^{50}}.3.5.7....99.101}} = {2^{50}}.\dfrac{1}{{{2^{50}}.101}} = \dfrac {1}{101}\\
b)\,\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{5} = t\\
\Rightarrow x = 2t;y = 3t;z = 5t\\
Nen\,xy + yz + xz - 31 = 0\\
\Rightarrow 2t.3t + 3t.5t + 2t.5t - 31 = 0\\
\Rightarrow 6{t^2} + 15{t^2} + 10{t^2} - 31 = 0\\
\Rightarrow 31{t^2} - 31 = 0\\
\Rightarrow t = 1\,\,hoac\,t = - 1\\
+ )\,t = 1 \Rightarrow x = 2;y = 3;z = 5\\
+ )\,t = - 1 \Rightarrow x = - 2;y = - 3;z = - 5
\end{array}\)
