A

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ =1-0-0-...-0-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

\(3x+6=0\Leftrightarrow3x=-6\Leftrightarrow x=-2\)

Vì 2≠-2 nên 2 pt không tương đương

Để điểm N(3;-5) đi qua đồ thị hàm số y=mx² thì

\(-5=m.3^2\\ \Rightarrow m.9=-5\\ \Rightarrow m=-\dfrac{5}{9}\)

Chọn D

Nửa chu vi hình chữ nhật là:
120:2=60(m)

Chiều dài hình chữ nhật là:

60:(2+1)x2=40(m)

Chiều rộng hình chữ nhật là:

60-40=20(m)

Đáp số:...

\(\dfrac{3}{13}+\dfrac{-3}{7}+\dfrac{10}{13}+\dfrac{-4}{7}\\ =\left(\dfrac{3}{13}+\dfrac{10}{13}\right)+\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\\ =\dfrac{13}{13}+\dfrac{-7}{7}\\ =1+\left(-1\right)\\ =0\)

\(a,3\left(x-1\right)-7=5\left(x+2\right)\\ \Leftrightarrow3x-3-7=5x+10\\ \Leftrightarrow3x-10=5x+10\\ \Leftrightarrow2x+20=0\\ \Leftrightarrow x=-10\\ b,\dfrac{3x+1}{2}-\dfrac{x+3}{5}=\dfrac{x}{10}+2\\ \Leftrightarrow\dfrac{5\left(3x+1\right)}{10}-\dfrac{2\left(x+3\right)}{10}-\dfrac{x}{10}-\dfrac{20}{10}\\ \Leftrightarrow15x+5-2x-6-x-20=0\\ \Leftrightarrow12x-21=0\\ \Leftrightarrow x=\dfrac{7}{4}\)

\(c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-2}{x+1}-\dfrac{x}{x-1}=\dfrac{x-8}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-8}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-x-x+8}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow-5x+10=0\\ \Leftrightarrow x=2\left(tm\right)\)

\(\left(x-2\right)\left(x^2-3x+5\right)=\left(2-x\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)=-\left(x-2\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)+\left(x-2\right)\left(1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5+1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-3x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\\ \Leftrightarrow x=2\)

an viết hoa