lúk trc hiện1 tn bây h nó hiện lên tận 2 tin

Nhắn tin vs thầy giáo thầy bảo thầy đag bận

52312 dug cho mk **** nhe

nhìn công thức ở đó mà làm

\(P=x^2+2y^2-2xy+8x+8y+2017\)

\(=\left(x^2-2xy+8x\right)+2y^2+8y+2017\)

\(=\left[x^2-2x\left(y-8\right)+\left(y-8\right)^2\right]+2y^2+8y+2017-y^2+16y-64\)\(=\left(x-y+8\right)^2+y^2+24y+1953\)

\(=\left(x-y+8\right)^2+\left(y^2+24y+144\right)+1809\)

\(=\left(x-y+8\right)^2+\left(y+12\right)^2+1809\ge1809\forall x\)Vậy Min P = 1809 khi \(\left\{{}\begin{matrix}x-y+8=0\\y+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+20=0\\y=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-12\end{matrix}\right.\)

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2=0\)\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2abxy-b^2y^2=0\)\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\)

\(\Rightarrow ay-bx=0\)

\(\Leftrightarrow ay=bx\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{y}{b}\)

=> đpcm

cj dùng Casio 570VN- Plus ^_^

Còn cái kia cj kh bk

Phân tích đa thức thành nhân tử:

\(a,x^3-6x^2+11x-6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

\(=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-5x+6\right)\)

\(=\left(x-1\right)\left(x^2-3x-2x+6\right)\)

\(=\left(x-1\right)\left[x\left(x-3\right)-2\left(x-2\right)\right]\)

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

\(D=x^2+2y^2-2xy+4x-3y-12\)

\(=\left(x^2-2xy+4x\right)+2y^2-3y-12\)

\(=\left[x^2-2x\left(y-2\right)+\left(y-2\right)^2\right]+2y^2-3y-12-y^2+4y-4\)\(=\left(x-y+2\right)^2+y^2+y+8\)

\(=\left(x-y+2\right)^2+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{31}{4}\)

\(=\left(x-y+2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\forall x\)

Vậy Min D = \(\dfrac{31}{4}\) khi \(\left\{{}\begin{matrix}x-y+2=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+\dfrac{5}{2}=0\\y=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)