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

\(A=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{y+z}{z}=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)

Mình giỏi đó h =)))

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{12.28}{12+28}=8,4\left(\Omega\right)\)

Do mắc nối tiếp nên \(I=I_1=1,8A\)

Do mắc song song nên \(U=U_1=U_2=I_1.R_1=1,8.12=21,6\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{21,6}{28}=\dfrac{27}{35}\left(A\right)\)

d) ĐKXĐ: \(x\ne5,x\ne-2\)

 \(pt\Leftrightarrow\dfrac{x^2-x-1}{\left(x-5\right)\left(x+2\right)}-\dfrac{1}{x-5}=0\Leftrightarrow\dfrac{x^2-x-1-\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-x-1-x-2=0\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

e) ĐKXĐ: \(x\ne1,x\ne2\)

\(pt\Leftrightarrow\dfrac{x+1}{x-1}+\dfrac{1}{x-2}-\dfrac{3x+4}{\left(x-1\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)+\left(x-1\right)-3x-4}{\left(x-1\right)\left(x-2\right)}=0\Leftrightarrow x^2-x-2+x-1-3x-4=0\Leftrightarrow x^2-3x-7=0\)

\(\Delta=\left(-3\right)^2-4.\left(-7\right)=37>0\)

PT có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{37}}{2}\left(tm\right)\\x_2=\dfrac{3+\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

1) ĐKXĐ: \(x\ne1,x\ne-1\)

2) \(A=\dfrac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+1}{2}\)

\(=\dfrac{4}{2\left(x-1\right)}=\dfrac{2}{x-1}\)

3) \(A=\dfrac{2}{x-1}=\dfrac{2}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}\)

4) \(A=\dfrac{2}{x-1}\in Z\Rightarrow\left(x-1\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp với ĐKXĐ \(\Rightarrow x\in\left\{0;2;3\right\}\)

Đặt \(S=1+5+5^2+5^3+...+5^{2016}\)

\(\Rightarrow5S=5+5^2+5^3+...+5^{2017}\)

\(\Rightarrow4S=5S-S=5+5^2+...+5^{2017}-1-5-...-5^{2016}=5^{2017}-1\)

\(\Rightarrow S=\dfrac{5^{2017}-1}{4}\)

Theo đề bài ta được: \(S.\left|x-1\right|=5^{2017}-1\)

\(\Leftrightarrow\dfrac{5^{2017}-1}{4}.\left|x-1\right|=5^{2017}-1\Leftrightarrow\dfrac{\left|x-1\right|}{4}=1\)

\(\Leftrightarrow\left|x-1\right|=4\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)