\(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-0,3\right)^5.\left(-8\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-8\right)^5}{\left(0.01\right)^3}=\dfrac{\left(-8.0,125\right)^5}{\left(0.01\right)^3}=\dfrac{-1}{0.000001}=-1000000\)

Lời giải:

câu a)

Lấy điểm $I,J$ thỏa mãn: \(\left\{\begin{matrix} 2\overrightarrow{IA}+\overrightarrow{IB}=\overrightarrow{0}\\ \overrightarrow{JA}+\overrightarrow{IB}+\overrightarrow{JC}=\overrightarrow{0}\end{matrix}\right.\)

Vì $A,B,C$ cố định nên $I,J$ cũng cố định.

Ta có:

\(|2\overrightarrow{MA}+\overrightarrow{MB}|=|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}|\)

\(\Leftrightarrow |2\overrightarrow{MI}+2\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}|=|\overrightarrow{MJ}+\overrightarrow{JA}+\overrightarrow{MJ}+\overrightarrow{JB}+\overrightarrow{MJ}+\overrightarrow{JC}|\)

\(\Leftrightarrow |3\overrightarrow {MI}|=|3\overrightarrow{MJ}|\Leftrightarrow |\overrightarrow{MI}|=|\overrightarrow{MJ}|\)

Do đó tập hợp điểm M nằm trên đường trung trực của \(IJ\)

câu b)

Lấy hai điểm $H,K$ thỏa mãn: \(\left\{\begin{matrix} 2\overrightarrow{HA}+\overrightarrow{HB}=\overrightarrow{0}\\ \overrightarrow{KA}+2\overrightarrow{KB}=\overrightarrow{0}\end{matrix}\right. \)

Vì $A,B,C$ cố định nên $H,K$ cũng cố định.

Ta có:

\(|2\overrightarrow{MA}+\overrightarrow{MB}|=|\overrightarrow{MA}+2\overrightarrow{MB}|\)

\(\Leftrightarrow |2\overrightarrow {MH}+2\overrightarrow{HA}+\overrightarrow{MH}+\overrightarrow{HB}|=|\overrightarrow{MK}+\overrightarrow{KA}+2\overrightarrow{MK}+2\overrightarrow{KB}|\)

\(\Leftrightarrow |3\overrightarrow{MH}|=|3\overrightarrow{MK}|\Leftrightarrow |\overrightarrow{MH}|=|\overrightarrow{MK}|\)

Do đó tập hợp điểm biểu diễn điểm $M$ nằm trên đường trung trực của $HK$

Có \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)=\(\left(\dfrac{-5}{4}+\dfrac{2}{4}\right)^2\)=\(\left(\dfrac{-3}{4}\right)^2\)=\(\dfrac{\left(-3\right)^2}{4^2}=\dfrac{9}{16}\)

Có \(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,125.2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,3\right)^5}{\left(-0.3\right)^5.\left(0,01\right)^3}=\dfrac{1}{-1.\left(0,01\right)^3}=\dfrac{1}{-\left(0,01\right)^3}\)

a/ \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}\)= \(\frac{625\times16}{3125\times\left(-32\right)}\)=\(\frac{10000}{-10000}\)= -1

\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=2^5\)

b) \(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{5^5.\left(-2\right)^5}=\dfrac{1.1}{5.\left(-2\right)}=\dfrac{1}{-10}=\dfrac{-1}{10}\)

( Mình giải được có 2 bài thôi. ) :)

a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=3^5\)

b)\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(25.4\right)^2}{\left(5.\left(-2\right)\right)^5}=\dfrac{\left(100\right)^2}{\left(-10\right)^5}=\dfrac{1}{-10}\)

Còn phần c mình cx đang tắc

\(\frac{\left(-0,125\right)^5\times\left(2,4\right)^5}{\left(-0,3\right)^5\times\left(0,01\right)^3}=\frac{\left(-0,12\times2,4\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{\left(-0,3\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{1}{0,000001}\)

\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)

\(=\dfrac{\left(0.125\cdot2.4\right)^5}{\left(-0.3\cdot0.01\right)^3\cdot\left(-0.3\right)^2}=\dfrac{0.3^5}{0.3^2\cdot\left(-0.003\right)^3}=\dfrac{0.3^3}{-0.003^3}=-1000000\)