Sửa đề: Cho hcn ABCD

\(\widehat{AEC}=\widehat{BAE}+\widehat{B}\\ =\dfrac{1}{2}\widehat{BAC}+\widehat{B}=\dfrac{1}{2}\left(\widehat{BAC}+\widehat{B}+\widehat{C}\right)+\dfrac{1}{2}\widehat{B}-\dfrac{1}{2}\widehat{C}\\ =\dfrac{1}{2}\cdot180^0+\dfrac{1}{2}\left(\widehat{B}-\widehat{C}\right)=90^0+\dfrac{1}{2}\cdot30^0=105^0\)

Xem lại đề

\(PT\Leftrightarrow\left(\dfrac{x-1}{2000}-1\right)+\left(\dfrac{x-2}{1999}-1\right)=\left(\dfrac{x-3}{1998}-1\right)+\left(\dfrac{x-4}{1997}-1\right)\\ \Leftrightarrow\dfrac{x-2001}{2000}+\dfrac{x-2001}{1999}=\dfrac{x-2001}{1998}+\dfrac{x-2001}{1997}\\ \Leftrightarrow\left(x-2001\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \Leftrightarrow x=2001\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\right)\)

\(S=10....00+8=10....08\) có tổng các chữ số là \(1+0+...+8=9⋮9\)

\(\Rightarrow S⋮9\)

Mà \(S\) có tận cùng 8 nên chia hết cho 2

Vậy \(S⋮2.9=18\)

\(\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\)

Bài 1:

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)

Bài 2:

\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)

Câu dưới: Do nhôm tác dụng đc với dd kiềm