=160- (2^3.5^2^3.5^2)
=160-0
=160 

\(160-\left(2^3.5^2-6.25\right)\)

\(=160-\left(8.25-6.25\right)\)

\(=160-\left[25.\left(8-6\right)\right]\)

\(=160-50\)

\(=110\)

A = 1.3+2.4+3.5 + ... + 99.101

<=> A= (2-1).(2+1)+(3-1).(3-1)+(4-1).(4+1)+...+(100-1).(100+1)

<=> A= 2² -1+3²-1+4²-1+....+100²-1

<=> A=(2²+3²+4²+...+100²)-(1+1+1+1+...+1)

<=>A=(2²+3²+4²+....+100²)-99

Và kết quả cuối cùng đó chính là 338250

Bài này vẫn còn 1 cách nữa nhưng cách đó dài quá nên mình làm hơi vắn tắt xíu

\(\Rightarrow Rtd=\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{Rbc.R1}{Rbc+R1}=\dfrac{12.\left(\dfrac{6^2}{6}\right)}{12+\dfrac{6^2}{6}}+\dfrac{12.12}{12+12}=10\Omega\)

\(b,\Rightarrow Ibc1=\dfrac{U}{Rtd}=\dfrac{12}{10}=1,2A\Rightarrow Ubc1=Ibc1\left(\dfrac{Rbc.R1}{Rbc+R1}\right)=7,2V\Rightarrow I1=\dfrac{7,2}{R1}=0,6A\Rightarrow Q1=I1^2R1t=1296W\)

\(c,\Rightarrow\left\{{}\begin{matrix}Ud=6V=Uac\\Id=\dfrac{Pdm}{Udm}=1A\end{matrix}\right.\)

\(\Rightarrow\)\(\dfrac{12}{Rtd}=\dfrac{12}{\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{\left(24-Rac\right)R1}{24-Rac+R1}}=\dfrac{12}{\dfrac{6Rac}{6+rac}+\dfrac{\left(24-Rac\right).12}{36-Rac}}=Iacd\)

\(\Rightarrow1+Iac=Iacd\Rightarrow1+\dfrac{6}{Rac}=\dfrac{12}{\dfrac{6Rac}{6+Rac}+\dfrac{\left(24-Rac\right)12}{36-Rac}}\Rightarrow Rac=12\sqrt{2}\left(\Omega\right)\)

3. R4 nt {R1//(R2ntR3)}

\(a,\Leftrightarrow\)\(Ia=0,3A=I2=I3\Rightarrow U23=U123=I2.\left(R2+R3\right)=6V\)

\(\Rightarrow Im=\dfrac{U123}{R123}=\dfrac{6}{\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}}=0,5A\Rightarrow Uab=Im.Rtd=0,5\left(R4+R123\right)=10V\)

\(b,\) R2//{R1 nt(R3//R4)}

\(\Rightarrow K\) mở \(\Rightarrow I3=\dfrac{U.R123}{Rtd.R23}=\dfrac{6}{12+R4}\left(A\right)\)

\(\Rightarrow K\) đóng \(\Rightarrow I3=\dfrac{U.R4}{R134.R34}=\dfrac{2R4}{30+7R4}\left(A\right)\)

\(\Rightarrow R4=15\Omega\)

\(\Rightarrow Ik=I2+I3=\dfrac{U}{R2}+\dfrac{2.15}{30+7.15}=\dfrac{10}{15}+\dfrac{2.15}{30+7.15}=\dfrac{8}{9}A\)

Đề bị lỗi hiển thị rồi bạn.