e,\(3\frac{2}{7}x-\frac{1}{8}=2\frac{3}{4}\)

\(=>\frac{23}{7}x-\frac{1}{8}=\frac{11}{4}\)

\(=>\frac{23}{7}x=\frac{11}{4}+\frac{1}{8}=\frac{23}{8}\)

\(=>x=\frac{23}{8}:\frac{23}{7}\)

\(=>x=\frac{7}{8}\)

b) 5 và 4/7 :x=13

  39/7 :x =13

          x= 39/7 :13

         x= 3/7

b) \(5\frac{1}{4}.\frac{3}{8}+10\frac{3}{4}.\frac{3}{8}\)

\(=\left(5\frac{1}{4}+10\frac{3}{4}\right).\frac{3}{8}\)

\(=16.\frac{3}{8}=6\)

c) \(6\frac{1}{5}.\frac{-2}{7}+14\frac{4}{5}.\frac{-2}{7}\)

\(=\left(6\frac{1}{5}+14\frac{4}{5}\right).\frac{-2}{7}\)

\(=21.\frac{-2}{7}=-6\)

đây nhá có gì thắc mắc hỏi chị nhá

3.

\(F=\dfrac{k.\left|q_1.q_2\right|}{r^2}=\dfrac{9.10^9.\left|9.10^{-18}\right|}{0,1^2}=8,1.10^{-6}N\)

1.

\(F=\dfrac{k.\left|q_1.q_2\right|}{r^2}=\dfrac{9.10^9.\left|5.10^8.\left(-1,6.10^{-19}\right).5.10^8.\left(-1,6.10^{-19}\right)\right|}{0,02^2}=1,44.10^{-7}N\)

Bài 1:

\(54\left(\dfrac{km}{h}\right)=15\left(\dfrac{m}{s}\right);9\left(\dfrac{m}{s}\right)=32,4\left(\dfrac{km}{h}\right)\)

Baì 2:

\(t'=s':v'=5:\left(5.3,6\right)=\dfrac{5}{18}h\)

\(\Rightarrow v_{tb}=\dfrac{s'+s''}{t'+t''}=\dfrac{5+3,8}{\dfrac{5}{18}+\left(\dfrac{15}{60}\right)}\simeq16,67\left(\dfrac{km}{h}\right)\)

Nhường e cái lý 8 với ạ :((

Câu 2: 

\(\Leftrightarrow\left(x+2\right)\left(10x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{3}{10}\end{matrix}\right.\)

\(A1+2=A2\)

   \(\overline{A}\) =\(\dfrac{A1\cdot54+\cdot\left(A1+2\right)\cdot46}{100}\)=79.92

\(\Leftrightarrow\)A1=79\(\Rightarrow\)A2=81  

Thanksss bann nhiều lắm nhoa