Tk quen đe🤣:

Tách nhỏ câu hỏi ra bạn

a) \(R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{4}}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\left(\Omega\right)\)

b) Do mắc song song nên \(U=U_1=U_2=U_3=12V\)

\(\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{12}{3}=4\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{12}{6}=2\left(A\right)\\I_3=\dfrac{U_3}{R_3}=\dfrac{12}{4}=3\left(A\right)\end{matrix}\right.\)

\(x^2-2\left(m-2\right)x+m^2+2m-3=0\left(1\right)\)

Để phương trình có hai nghiệm phân biệt thì Δ' > 0

\(\Rightarrow\left(m-2\right)^2-m^2-2m+3>0\Leftrightarrow m^2-4m+4-m^2-2m+3>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \dfrac{7}{6}\)\)

Theo viét : \(\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)\)

Lại có :\( \dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{5}\)

\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1+x_2}{5}\)

\(\Rightarrow\left(x_1+x_2\right)\left(x_1x_2\right)=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(2m-4\right)\left(m^2+2m-3\right)=5\left(2m-4\right)\)

\(\Leftrightarrow2m^3+4m^2-6m-4m^2-8m+12=10m-20\)

\(\Leftrightarrow2m^3-24m+32=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=-4\left(n\right)\\m=2\left(l\right)\end{matrix}\right.\)

Vậy \(m=-4\) thì thỏa điều kiện

a: Xét ΔAHC và ΔDHB có 

AH=DH

\(\widehat{AHC}=\widehat{DHB}\)

HC=HB

Do đó: ΔAHC=ΔDHB

Sai

mình ghi thiếu thông cảm nha

\(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)

\(\Rightarrow2\cdot3^x=10\cdot3^{12}+8\cdot\left(3^3\right)^4\)

\(\Rightarrow2\cdot3^x=10\cdot3^{12}+8\cdot3^{12}\)

\(\Rightarrow2\cdot3^x=18\cdot3^{12}\)

\(\Rightarrow3^x=\dfrac{18\cdot3^{12}}{2}\)

\(\Rightarrow3^x=9\cdot3^{12}\)

\(\Rightarrow3^x=3^2\cdot3^{12}\)

\(\Rightarrow3^x=3^{14}\)

\(\Rightarrow x=14\)

(1-5)

1)Light snack

2)reception

3)key

4)shopping

5)viewing

Xin lỗi vì ko làm được câu 6-10 vì bài thiếu thông tin :(