Bài 1:

\(R=U:I=15:0,1=150\Omega\)

\(\Rightarrow R_b=R-R'=150-50=100\Omega\)

\(I=I'=I_b=0,1A\left(R'ntR_b\right)\)

\(\left\{{}\begin{matrix}U'=R'.I'=50.0,1=5V\\U_b=R_b.I_b=100.0,1=10V\end{matrix}\right.\)

\(I_{sau}=I+0,15=0,1+0,15=0,25A\)

\(\Rightarrow R_{bsau}=U_b:I_{sau}=10:0,25=40\Omega\)

Bài 2:

\(R=U:I=25:1=25\Omega\)

\(\Rightarrow\dfrac{1}{R_b}=\dfrac{1}{R}-\dfrac{1}{R'}=\dfrac{1}{25}-\dfrac{1}{50}=\dfrac{1}{50}\Rightarrow R_b=50\Omega\)

\(U=U'=U_b=25V\)(R'//R_b)

\(\left\{{}\begin{matrix}I'=U':R'=25:50=0,5A\\I_b=U_b:R_b=25:50=0,5A\end{matrix}\right.\)

4:

a: 7/10

b: 6/5=12/10

7/2=35/10

8/25=32/100

11/50=22/100

23/125=204/1000

17/20=85/100

41/200=205/1000

Ta có:                                  \(A=\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3+...+\left(-7\right)^{200}\)
  \(\Rightarrow\)                     \(\left(-7\right)A=\left(-7\right)^2+\left(-7\right)^3+\left(-7\right)^4+...+\left(-7\right)^{201}\)
  \(\Rightarrow\)\(A-\left(-7\right)A=8A=\left(-7\right)-\left(-7\right)^{201}\)
  \(\Rightarrow\)                                \(A=\frac{\left(-7\right)-\left(-7\right)^{201}}{8}=\frac{\left(-7\right)+7^{201}}{8}\)

A=(-7)+(-7)^2+...+(-7)^200

7a=-[7^2+7^3+...+7^201]

7a-a=-[(7^2+7^3+...+7^201)-(7+7^2+...+7^200)]

6a=-(7^2+7^3+...+7^201-7-7^2+...+7^200)

6a=-(7^201-7)

a=-\(\frac{-\left(7^{201}-7\right)}{6}\)

\(\left(x-\frac{3}{5}\right)=\frac{2}{5}×-\frac{1}{3}\)

\(\left(x-\frac{3}{5}\right)=-\frac{2}{165}\)

\(x=-\frac{2}{165}+\frac{3}{5}\)

\(x=\frac{97}{165}\)

vậy \(x=\frac{97}{165}\)

\(x×\left(\frac{3}{7}+\frac{2}{3}\right)=\frac{10}{21}\)

\(x×\frac{23}{21}=\frac{10}{21}\)

\(x=\frac{10}{21}:\frac{23}{21}\)

\(x=\frac{10}{23}\)

vậy \(x=\frac{10}{23}\)

\(\left(x-\frac{3}{5}\right):\frac{-1}{3}=\frac{2}{5}\)

=> \(x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)

=> \(x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}=\frac{7}{15}\)

\(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)

=> \(\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)

=> \(-\frac{5}{21}x=\frac{10}{21}\)

=> \(x=\frac{10}{21}:\frac{-5}{21}=\frac{10}{21}\cdot\frac{-21}{5}=-2\)

Hai bài của ☆luffy cute☆ đều sai hết , xem xét lại đi nhé

1 ) \(\left(x-\frac{3}{5}\right)\div-\frac{1}{3}=\frac{2}{5}\)

\(\Rightarrow x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)

\(\Rightarrow x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}\)

\(\Rightarrow x=\frac{7}{15}\)

2 ) \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)

\(\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)

\(\Rightarrow-\frac{5}{21}x=\frac{10}{21}\)

\(\Rightarrow x=\frac{10}{21}\div-\frac{5}{21}\)

\(\Rightarrow x=-2\)

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b cần bài nào thế

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