\(x^4+2000x^2+1999x+2000\)

\(=\left(x^4-x\right)+\left(2000x^2+2000x+2000\right)\)

\(=x\left(x^3-1\right)+2000\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2000\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2000\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+2000\right)\)

a) Xét hình thang ABCD, có: AB // CD (giả thiết)

\(\Rightarrow\widehat{BAD}+\widehat{ADC}=180^o\) (trong cùng phía bù nhau)

Mà \(\widehat{BAD}=\widehat{ADC}\) (giả thiết)

\(\Rightarrow\widehat{BAD}=\widehat{ADC}=\dfrac{180^o}{2}=90^o\)

Xét hình thang ABCD (AB // CD), có: \(\widehat{BAD}=\widehat{ADC}=90^o\) (chứng minh trên)

=> ABCD là hình thang vuông (điều phải chứng minh)

b) Áp dụng định lí Pytago vào \(\Delta ACD\) vuông tại A, có: \(AC^2=AD^2+CD^2\)

Áp dụng định lí Pytago vào \(\Delta ABD\) vuông tại A, có: \(BD^2=AB^2+AD^2\)

Do đó: \(AC^2+BD^2=\left(AD^2+CD^2\right)+\left(AB^2+AD^2\right)=AB^2+CD^2+2AD^2\) (điều phải chứng minh)

1. Rice is grown in tropical countries.
2. His children are loved very much by Mr. Green.
3. French isn't spoken here.
4. His windows were broken last night.
5. Old men should be treated with respect by children.
6. All the accounts can be done by the computer.

/i:/: convenient, beach, street, police, cheap, city, sleepy
/i/: friendly, river, village, exciting, sit

I can't remember a suggestion of the trip was

Yêu cầu bài toán \(\Leftrightarrow\left\{{}\begin{matrix}m^2-4=5\\2m-1\ne3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=9\\m\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=3\\m=-3\end{matrix}\right.\\m\ne-3\end{matrix}\right.\Leftrightarrow m=3\)

Vậy với \(m=3\) thì hai hàm số trên song song.

a) \(A=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)

\(=\left(x^2-3x\right)^2-1\ge-1\)

Dấu bằng xảy ra \(\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy \(MinA=-1\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

b) \(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)

\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)

\(=\left(x^2+4x\right)^2-25\)

Dấu bằng xảy ra \(\Leftrightarrow x^2+4x=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(MinB=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

\(D=\dfrac{1}{1-\dfrac{1}{1-2^2}}+\dfrac{1}{1+\dfrac{1}{1+2^2}}\)

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

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

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

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

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

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

\(=2-\left[\dfrac{1}{2^2}+\dfrac{1}{2\left(2+1\right)}\right]\)

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

\(=2-\dfrac{5}{2^2.3}\)

\(=\dfrac{2^3.3-5}{2^2.3}\)

\(=\dfrac{8.3-5}{4.3}\)

\(=\dfrac{24-5}{12}\)

\(=\dfrac{19}{12}\)

\(-3n^2-3=-3n^2+\left(-3\right).1=-3\left(n^2+1\right)\)

\(\dfrac{y}{15}< \dfrac{4}{45}\)

\(\dfrac{y\times3}{15\times3}< \dfrac{4}{45}\)

\(\dfrac{y\times3}{45}< \dfrac{4}{45}\)

\(\Rightarrow y\times3< 4\)

\(y< \dfrac{4}{3}\)

\(\dfrac{25}{13}< \dfrac{5}{y}\)

\(\dfrac{25}{13}< \dfrac{5\times5}{y\times5}\)

\(\dfrac{25}{13}< \dfrac{25}{y\times5}\)

\(\Rightarrow y\times5< 13\)

\(y< \dfrac{13}{5}\)