a: \(A=x^2-2x+1+y^2-4y+4+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=2

b: \(B=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x=0 hoặc x=-5

a) \(A=x^2-2x+y^2-4y+7\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(A_{min}=2\Leftrightarrow x=1;y=2\).

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

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

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

-Đặt \(t=x^2+5x-6\)

\(B=t\left(t+12\right)=t^2+12t=t^2+12t+36-36=\left(t+6\right)^2-36\ge-36\)

\(B_{min}=-36\Leftrightarrow t=-6\Leftrightarrow x^2+5x-6=-6\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow x=0hayx=-5\)

\(A=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)

\(A_{min}=10\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

\(B=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)

\(B_{min}=-36\) khi \(x^2+5x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(C=\left(x^2-2x+1\right)+\left(y^2-4x+4\right)+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(1;2\right)\)

a. \(A=4x^2+4x+11\)

   \(A=\left(4x^2+4x+1\right)+10\)

  \(A=\left(2x+1\right)^2+10\)

Ta có: \(\left(2x+1\right)^2\ge0;\forall x\) 

\(\Rightarrow A_{min}=10\)

Dấu "=" xảy ra khi \(\left(2x+1\right)^2=0\)

                            \(\Leftrightarrow2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)

c.\(C=x^2-2x+y^2-4y+7\)

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

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

Ta có: \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0;\forall x,y\)

\(\Rightarrow C_{min}=2\)

Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

3. Tìm giá trị nhỏ nhất của các biểu thứca. A = 4x2  4x 11b. B = (x - 1) (x 2) (x 3) (x 6)c. C = x2 - 2x y2 - 4y 7Ai nha... - Hoc24

Yêu cầu đề là gì vậy bạn?

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

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

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy \(S=\left\{2;2\right\}\)

Lần sau bạn chú ý viết đầy đủ yêu cầu của đề bài.

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

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

c) \(C=9x^2-6x+3=3\left(3x^2-2x+1\right)\)

nè bạn Câu hỏi của Hương Linh - Toán lớp 8 - Học toán với OnlineMath

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

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

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

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

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)