Bài 1:

\(\Delta ABC\) vuông tại A, theo định lí Py-ta-go:

Ta có: \(BC^2=AB^2+AC^2\)

\(\Rightarrow AB^2=BC^2-AC^2\)

\(AB^2=10^2-6^2\)

\(AB^2=64\)

\(\Rightarrow AB=\sqrt{64}=8\left(cm\right)\).

Ta có: x = 2011 \(\Rightarrow\) 2010 = x - 1

\(A=x^{2011}-2010x^{2010}-2010x^{2009}-...-2010x+1\)

\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)

\(=x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)

\(=x^{2011}-x^{2011}+x^{2010}-x^{2010}+x^{2009}-...-x^2+x+1\)

\(=x+1\)

\(=2011+1\)

\(=2012.\)

Ta có: MN là đường trung bình của \(\Delta ABH\) (vì M, N lần lượt là trung điểm của AH, BH)

\(\Rightarrow\) MN // AB và MN = \(\dfrac{1}{2}\)AB

Mà AB // CD và AB = CD

\(\Rightarrow\) MN // CD và MN = \(\dfrac{1}{2}\)CD

hay MN // CK và MN = CK

\(\Rightarrow\) MNCK là hình bình hành

\(\Rightarrow\) MK // NC (1)

Lại có: \(\left\{{}\begin{matrix}MN//AB\left(cmt\right)\\AB\perp BC\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow\) \(MN\perp BC\) tại E (E \(\in\) BC)

\(\Delta BCM\) có hai đường cao BH và ME cắt nhau tại N

\(\Rightarrow CN\perp BM\) (2)

Từ (1) và (2) \(\Rightarrow BM\perp MK\) (đpcm).

\(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)

\(=\dfrac{2x+6}{3x^2-x}.\dfrac{1-3x}{x^2+3x}\)

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

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

\(=-\dfrac{2}{x^2}.\)

mẹ 35 tuổi

con 5 tuổi

a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)

\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)

\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)

\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)

b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)

\(=4x^2-16.\dfrac{7x-2}{3x+6}\)

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

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

\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)

c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)

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

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

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

\(=\dfrac{3\left(x+1\right)}{x-1}\)

d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)

\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)

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

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

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