HF x ??

Xét \(\Delta\text{A}BC\)có :

 \(ED//\text{A}C\left(gt\right)\)

\(\Rightarrow\frac{BE}{\text{A}B}=\frac{DE}{\text{A}C}\)

\(\Rightarrow\frac{BE}{ED}=\frac{\text{A}B}{\text{A}C}(1)\)

Có : AD là phân giác góc \(B\text{A}C\)

=> góc \(B\text{A}D\)=  góc \(C\text{A}D\) 

Có : \(ED//\text{A}C\left(gt\right)\)

=> góc \(\text{A}DE\)=  góc \(C\text{A}D\) 

mà  góc \(B\text{A}D\)=  góc \(C\text{A}D\) ( cmt)

=> góc \(\text{A}DE\)=  góc \(B\text{A}D\)

=> \(\Delta ED\text{A}\) cân tại E

=> \(ED=E\text{A}\)

Cộng mỗi vế của (1) với 1, ta có : 

\(1+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+1\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+\frac{ED}{ED}\)

mà \(ED=E\text{A}\left(cmt\right)\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+\frac{E\text{A}}{ED}\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{\text{A}B}{ED}\)

=>\(\frac{1}{\text{A}B}+\frac{1}{\text{A}C}=\frac{1}{ED}\)

mà  ​​\(ED=E\text{A}\left(cmt\right)\)

=> \(\frac{1}{\text{A}B}+\frac{1}{\text{A}C}=\frac{1}{E\text{A}}\left(đpcm\right)\)

Không mất tính tổng quát, giả sử \(x^2\ge y^2\)

suy ra \(1=x^2+y^2-xy\ge y^2\)

\(\Rightarrow y\in\left\{-1,0,1\right\}\)(vì \(y\)nguyên) 

Với \(y=-1\): \(x^2+1+x=1\Leftrightarrow x^2+x=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)(thỏa mãn) 

Với \(y=0\): \(x^2=1\Leftrightarrow x=\pm1\)

Với \(y=1\): \(x^2+1-x=1\Leftrightarrow x^2-x=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)(thỏa mãn) 

Câu 12: 

 \(x^2-4x+y^2-6y+15=2\)

\(\Leftrightarrow x^2-4x+4+y^2-6y+9=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

ko có biết

\(Q=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Ta có: 

\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)

Tương tự: 

\(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)

\(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)

Cộng lại vế với vế ta được: 

\(\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y-z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}=0\)

Suy ra \(2Q=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\ge\frac{\frac{\left(x^2+y^2\right)^2}{2}}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{\frac{\left(z^2+x^2\right)^2}{2}}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)

\(\ge\frac{\frac{\left(x+y\right)^2}{2}}{2\left(x+y\right)}+\frac{\frac{\left(y+z\right)^2}{2}}{2\left(y+z\right)}+\frac{\frac{\left(z+x\right)^2}{2}}{2\left(z+x\right)}\)

\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}\)

\(=\frac{x+y+z}{2}=\frac{1}{2}\)

Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{3}\).

Vậy \(minQ=\frac{1}{4}\)đạt tại \(x=y=z=\frac{1}{3}\).

\(\Leftrightarrow\left(a.a+a.(−2)−1a−1.−2)(a−3)(a−1\right)\)

\(\Leftrightarrow\left(a^2-3a+2\right)\left(a-3\right)\left(a-1\right)\)

\(\Leftrightarrow\left(a^3-6a^2+11a-6\right)\left(a-1\right)\)

\(\Leftrightarrow a^4-7a^3+17a^2-17a+6\)