1 Tính: a $\frac{25^5.5^5}{65^10}$ b $\frac{15^3 +5.15^2-5^3}{18^3 +6.18^2-6^3}$ 2Tìm x: (x- $\frac{1}{3}$ ). (x+$\frac{2}{3}$ )$\geq$ 0

0 bình luận về “1 Tính: a $\frac{25^5.5^5}{65^10}$ b $\frac{15^3 +5.15^2-5^3}{18^3 +6.18^2-6^3}$ 2Tìm x: (x- $\frac{1}{3}$ ). (x+$\frac{2}{3}$ )$\geq$ 0”

1. Đáp án: 

   1) a) $\frac{5^5}{13^{10}}$

       b) $\frac{125}{216}$

   2) \(\left[ \begin{array}{l}x≥\frac{1}{3}\\x≤\frac{-2}{3}\end{array} \right.\) 

   Giải thích các bước giải:

   1) a) $\frac{25^5.5^5}{65^{10}}$

   =$\frac{(5^2)^5.5^5}{(5.13)^{10}}$ 

   =$\frac{5^{10}.5^5}{5^{10}.13^{10}}$ 

   =$\frac{5^5}{13^{10}}$ 

       b) $\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}$

   = $\frac{(3.5)^3+5.(3.5)^2-5^3}{(6.3)^3+6.(6.3)^2-6^3}$

   = $\frac{3^3.5^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}$

   = $\frac{5^3(3^3+3^2-1)}{6^3(3^3+3^2-1)}$

   = $\frac{5^3}{6^3}$

   = $\frac{125}{216}$

   2) $(x-\frac{1}{3})(x+\frac{2}{3})≥0$ 

   ⇔ $\left[ \begin{array}{l}\left \{ {{x-\frac{1}{3}≥0} \atop {x+\frac{2}{3}≥0}} \right.\\\left \{ {{x-\frac{1}{3}≥0} \atop {x+\frac{2}{3}}≥0} \right.\end{array} \right.$

   ⇔ $\left[ \begin{array}{l}\left \{ {{x≥\frac{1}{3}} \atop {x≥\frac{-2}{3}}} \right.\\\left \{ {{x≤\frac{1}{3}} \atop {x≤\frac{-2}{3}}} \right.\end{array} \right.$

   ⇔ \(\left[ \begin{array}{l}x≥\frac{1}{3}\\x≤\frac{-2}{3}\end{array} \right.\) 

   Trả lời

2. Bài 1:
   $a) \dfrac{25^{5}.5^{5}}{65^{10}}\\=\dfrac{5^{10}.5^{5}}{13^{10}.5^{10}}\\=\dfrac{5^{10}}{13^{10}.5^{5}}\\=\dfrac{5^{5}}{13^{10}}=\dfrac{3125}{23^{10}}\\b)\dfrac{15^{3}+5.15^{2}-5^{3}}{18^{3}+6.18^{2}-6^{3}}\\=\dfrac{3375+5.15^{2}-125}{5832+6.18^{2}-216}\\=\dfrac{3250+5.225}{5616+6.324}\\=\dfrac{3250+1125}{5616+1944}\\=\dfrac{4375}{7560}$
   Bài 2:
   $\left ( x-\dfrac{1}{3} \right )\left ( x+\dfrac{2}{3} \right )\geq 0\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
   x-\dfrac{1}{3}\geq 0\\ 
   x+\dfrac{2}{3}\geq 0
   \end{matrix}\right.\\\left\{\begin{matrix}
   x-\dfrac{1}{3}\leq 0\\ 
   x+\dfrac{2}{3}\leq 0
   \end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
   x\geq \dfrac{1}{3}\\ 
   x\geq -\dfrac{2}{3}
   \end{matrix}\right.\\\left\{\begin{matrix}
   x\leq \dfrac{1}{3}\\ 
   x\leq -\dfrac{2}{3}
   \end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\geq \dfrac{1}{3}\\x\leq -\dfrac{2}{3}\end{array} \right.$

    

   Trả lời