ĐKXĐ: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\2\sqrt{x}+3\left(3\sqrt{x}-5\right)=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\11\sqrt{x}=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\\sqrt{x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x^2+y^2=5\\x^2-3y^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\left(3y^2+1\right)+y^2=5\\x^2=3y^2+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10y^2=2\\x^2=3y^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{1}{5}\\x^2=\dfrac{8}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{1}{\sqrt{5}}\\y=\pm\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2-6x^2+9y^2-14y^2=108-74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\-5y^2=34\end{matrix}\right.\)

Vì \(-5y^2=34\Rightarrow y^2=\dfrac{34}{-5}< 0\) vô nghiệm

Vậy hệ phương trình đã cho vô nghiệm

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\5y^2=-34\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\y^2=-\dfrac{34}{5}< 0\left(vô-lý\right)\end{matrix}\right.\)

Vậy hệ pt vô nghiệm

\(\left\{{}\begin{matrix}7x^2+13y=-39\\5x^2-11y=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}77x^2+143y=-429\\65x^2-143y=429\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}143x^2=0\\65x^2-143y=429\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\)

ĐKXĐ: \(x\ne3;y\ne-1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-3}=u\\\dfrac{1}{y+1}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4u+5v=2\\5u+v=\dfrac{29}{20}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4u+5\left(\dfrac{29}{20}-5u\right)=2\\v=\dfrac{29}{20}-5u\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-21u=-\dfrac{21}{4}\\v=\dfrac{29}{20}-5u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{1}{4}\\v=\dfrac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-3}=\dfrac{1}{4}\\\dfrac{1}{y+1}=\dfrac{1}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=4\end{matrix}\right.\)

a) Vì vôn kế đo đoạn mạch R₁ => U₁ =5,4 (V)

I₁ = U₁/R₁ =5,4/8 =0,675 (A)

Vì R₁ nt R₂ => I₁= I₂= I =0,675 (A) 

b) R₁ nt R₂ => R_tđ =R₁ +R₂ =8+36 =44 (Ω)

U_tđ  = I.R_tđ =0,675.44 =29,7 (A)

\(a,\) Đối với hai điện trỏ mắc nối tiếp : \(R_{Tđ}=R_1+R_2\)

Đối với hai điện trỏ mắc song song 

  \(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\)

hay \(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}\)

\(b,\)  TH1 : Mắc nối tiếp 

\(R_{tđ}=R_1+R_2=40+60=100\left(\Omega\right)\)

TH2: Mắc song song

\(R_{Tđ}=\dfrac{R_1.R_2}{R_1+R_2}\\ =\dfrac{40.60}{40+60}=24\left(\Omega\right)\)

a) +) mắc nối tiếp: R_tđ =R₁ +R₂

+) mắc song song: 1/R_tđ =1/R₁ +1/R₂

b) TH1:mắc nối tiếp

 R_{tđ =}R₁ +R₂ = 40 +60 =100 (Ω)

TH2:mắc song song

1/R_tđ =1/R₁ +1/R₂ =1/40 +1/60 =1/24

=> R_tđ =24 (Ω)