\(A=\dfrac{\left(1+2+2+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(75,6-75,6\right)}{1-2+3-4+...+99-100}\)

\(A=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{1-2+3-4+...+99-100}=0\)

\(\left|2x-5\right|+\left|2x+3\right|=8\)

Mặt khác:\(\left|2x-5\right|+\left|2x+3\right|=\left|2x-5\right|+\left|-2x-3\right|\ge\left|2x-5-2x-3\right|=\left|-8\right|=8\)

Nên để thõa mãn yêu cầu thì:

\(\left(2x-5\right)\left(-2x-3\right)\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\-2x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\-2x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow-\dfrac{3}{2}\le x\le\dfrac{5}{2}\)

Cái này làm r mà ??

Sửa lại ĐKXĐ là \(x\ne\pm y\) nha

ĐKXĐ: \(x\ne y\)

a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)

b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)

\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)

a=7

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{12.18}{12+18}=7,2\left(\Omega\right)\)

\(U=U_1=U_2=I_1.R_1=0,75.12=9\left(V\right)\)

\(I=\dfrac{U}{R_{tđ}}=\dfrac{9}{7,2}=1,25\left(A\right)\)

\(A=\dfrac{x+y}{2\left(x+y\right)}\left(đk:x+y\ne0\right)\)

Vậy với \(x+y=0\) thì \(A\in\varnothing\)

\(\dfrac{R_1}{R_2}=\dfrac{S_2}{S_1}\Rightarrow R_2=\dfrac{R_1.S_1}{S_2}=\dfrac{8,5.5}{0,5}=85\left(\Omega\right)\)

\(A=P.t=U.I.t\Rightarrow I=\dfrac{A}{U.t}=\dfrac{1800}{120.60}=0,25\left(A\right)\)