cho B=

a/rút gọn B

b/tìm x để B=3

\(\frac{\sqrt[3]{20+14\sqrt{2}}}{\sqrt[3]{20-14\sqrt{2}}}\)

tìm x

\(\sqrt{x}+\sqrt{x-7}+\sqrt{x-12}=9\)

Rút gọn

A=\(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

B=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+\sqrt{2\sqrt{5}}}}\)

Vì R1//R2 nên

U=U1=U2

Mà U=Rtđ*I=\(\frac{R1\cdot R2}{R1+R2}\cdot I12\)

⇒U1=U2=\(\frac{R1\cdot R2}{R1+R2}\cdot I12\)

Ta có I1=\(\frac{U1}{R1}=\frac{\frac{R1\cdot R2}{R1+R2}\cdot I12}{R1}=\frac{\frac{R1\cdot R2}{R1+R2}}{R1}\cdot I12=\frac{R2}{R1+R2}\cdot I12\left(đpcm\right)\)

U1=R1*I1=35*2,4=84V

Vì R1 //R2 ta có

U=U2=U1=84V

I=U\(\frac{U}{Rtđ}=\frac{84}{\frac{35\cdot60}{35+60}}=3,8\left(A\right)\)

a)\(I1=\frac{U1}{R1}=\frac{15}{10}=1,5\left(A\right)\)

Vì R1 nt R2 nên

I=I2=I1=1,5A

U=I*Rtđ=I*(R1+R2)=1,5*40=60V

b)I'=\(\frac{I}{2}=\frac{1,5}{2}=0,75\left(A\right)\)

Rtđ'=\(\frac{U}{I'}=\frac{60}{0,75}=80\)Ω

Vì R1 nt R2 nt R3 nên

Rtđ'=R1+R2+R3

⇒80=10+20+R3

⇒R3=50Ω

VÌ R1//R2//R3 nên

U1=U2=U3=U=60V

\(\frac{1}{Rtđ}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\)

⇒\(\frac{1}{\frac{60}{9}}=\frac{1}{R1}+\frac{1}{\frac{R1}{2}}+\frac{1}{\frac{R1}{3}}\)

⇒\(\frac{3}{20}=\frac{1}{R1}+\frac{2}{R1}+\frac{3}{R1}\)

⇒\(\frac{3}{20}=\frac{1+2+3}{R1}\)

⇒R1=40Ω

⇒R2=\(\frac{R1}{2}\)=20Ω

⇒R3=\(\frac{R1}{3}\)=\(\frac{40}{3}\)Ω

I1=\(\frac{U1}{R1}=\frac{60}{40}=1,5\)(A)

I2=\(\frac{U2}{R2}=\frac{60}{20}=3\left(A\right)\)

I3=I-I1-I2=9-1,5-3=4,5(A)

Vì R1 // R2//R3