Bài 11:

\(PTHH:2A+Cl_2\rightarrow2ACl\\TheoĐLBTKL:\\ m_A+m_{Cl_2}=m_{ACl}\\ \Leftrightarrow 9,2+m_{Cl_2}=23,4\\ \Rightarrow m_{Cl_2}=23,4-9,2=14,2\left(g\right)\\ n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_A=2.0,2=0,4\left(mol\right)\\ M_A=\dfrac{9,2}{0,4}=23\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\left(I\right):Natri\left(Na=23\right)\)

8.31:

a: Xét ΔABD có AM/AB=AQ/AD

nên MQ//BD và MQ=BD/2

Xét ΔCBD có CN/CB=CP/CD

nên NP//BD và NP=BD/2

=>MQ//NP và MQ=NP

XétΔBAC có BM/BA=BN/BC

nên MN//AC

=>MN vuông góc BD

=>MN vuông góc MQ

Xét tứ giác MNPQ có

MQ//NP

MQ=NP

góc NMQ=90 độ

=>MNPQ là hình chữ nhật

=>M,N,P,Q cùng nằm trên 1 đường tròn

Bạn học trường nào ?

quan trung cấp 2

Bài 1:

a, \(\dfrac{2}{3}\) + \(\dfrac{1}{5}\). \(\dfrac{10}{7}\)

= \(\dfrac{2}{3}\) + \(\dfrac{2}{7}\) 

= \(\dfrac{20}{21}\)

b, \(\dfrac{7}{12}\) - \(\dfrac{27}{7}\). \(\dfrac{1}{18}\)

= \(\dfrac{7}{12}\) - \(\dfrac{3}{14}\)

= \(\dfrac{31}{84}\)

c, \(\dfrac{3}{10}\). \(\dfrac{-5}{6}\) - \(\dfrac{1}{8}\)

= - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\)

= - \(\dfrac{3}{8}\)

d, - \(\dfrac{4}{9}\): \(\dfrac{8}{3}\) + \(\dfrac{1}{18}\)

= - \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\)

= - \(\dfrac{1}{9}\)

e,  {[(\(\dfrac{1}{2}\) - \(\dfrac{2}{3}\))² : 2 ] - 1}. \(\dfrac{4}{5}\)

= {[ (-\(\dfrac{1}{6}\))² : 2] - 1}. \(\dfrac{4}{5}\)

= { [\(\dfrac{1}{36}\) : 2] - 1}. \(\dfrac{4}{5}\)

= { \(\dfrac{1}{72}\) - 1}. \(\dfrac{4}{5}\)

=- \(\dfrac{71}{72}\).\(\dfrac{4}{5}\)

= -\(\dfrac{71}{90}\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(a^3+b^3+c^3=3abc\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(a^2+b^2+c^2-ab-ac-bc=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a^2-2ba+b^2\right)+\left(b^2-2cb+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow a=b=c\)

\(A=\dfrac{a^{2023}}{b^{2023}}+\dfrac{b^{2023}}{c^{2023}}+\dfrac{c^{2023}}{a^{2023}}\)

\(=\dfrac{a^{2023}}{a^{2023}}+\dfrac{b^{2023}}{b^{2023}}+\dfrac{c^{2023}}{c^{2023}}\)

=1+1+1

=3