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

\(\Leftrightarrow4x^2-4x+1+y^2+2y+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y+1\right)^2=0\)

Đẳng thức xảy ra khi x = 1/2 ; y = - 1

Trả lời:

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

\(\Leftrightarrow4x^2+y^2-4x+2y+1+1=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}}\)

Vậy x = 1/2; x = - 1 là nghiệm của pt.

\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+2b+b^2=2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2-2ab=0\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a=b\).

\(A=5x^2-2x+7=5\left(x^2-\frac{2}{5}x+\frac{1}{25}-\frac{1}{25}\right)+7\)

\(=5\left(x-\frac{1}{5}\right)^2-\frac{1}{5}+7=5\left(x-\frac{1}{5}\right)^2+\frac{34}{5}\ge\frac{34}{5}\)

Dấu ''='' xảy ra khi x = 1/5

Vậy GTNN của A bằng 34/5 tại x = 1/5

\(C=x\left(x-1\right)\left(x-2\right)\left(x-3\right)+10\)

\(=\left(x^2-3x\right)\left(x^2-3x+2\right)+10\)

Đặt \(x^2-3x=t\)

\(t\left(t+2\right)+10=t^2+2t+10=t^2+2t+1+9=\left(t+1\right)^2+9\ge9\)

Dấu ''='' xảy ra khi \(x^2-3x+1=0\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\)

Vậy GTNN của C bằng 9 tại x = \(\frac{3\pm\sqrt{5}}{2}\)

\(VT=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}.\left(a+b+c\right)\)

\(VT=\frac{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2}{2}.\left(a+b+c\right)\)

\(VT=\frac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}.\left(a+b+c\right)\)

\(VT=\frac{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}{2}.\left(a+b+c\right)\)

\(VT=\left(a^2+b^2+c^2-ab-bc-ca\right).\left(a+b+c\right)\)

\(VT=a^3+b^3+c^3-3abc=VP\left(đpcm\right)\)

Đề bài là j vậy bn

bruhbruh

Trả lời:

\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\) \(\left(ĐK:a\ne b\ne c\right)\)

\(=\frac{1}{\left(a-b\right)\left(a-c\right)}-\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{b-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

Bạn thấy ảnh ko ạ ?

* Nguồn : Lazi *

Trả lời:

\(\left(ab-1\right)^2+\left(a+b\right)^2\)

\(=a^2b^2-2ab+1+a^2+2ab+b^2\)

\(=a^2b^2+1+a^2+b^2\)

\(=\left(a^2b^2+a^2\right)+\left(b^2+1\right)\)

\(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)

\(=\left(b^2+1\right)\left(a^1+1\right)\)