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12 tháng 9 2021

a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)

c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)

\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)

d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)

\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)

e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)

\(minE=-20\Leftrightarrow x=-3\)

f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)

\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)

12 tháng 9 2021

\(A=x^2+3x+4=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)

Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Mấy câu còn lại làm tương tự nhé em^^

a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)

\(=x^2-2xy+y^2+4y^2+4y+1+50\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

27 tháng 9 2021

a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)

\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)

c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)

\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)

d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)

\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)

\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)

25 tháng 10 2019

\(A=x^2+4x+100\)

\(A=x^2+2.x.2+2^2+96\)

\(A=\left(x+2\right)^2+96\)

           \(\left(x+2\right)^2+96\le0\)

           \(\left(x+2\right)^2+96\le96\)

    \(\Leftrightarrow A\le96\)

\(A_{min}\Leftrightarrow A=10\)

Dấu "=" xảy ra : \(\left(x+2\right)^20\)

                             \(x+2=0\)

                             \(x=-2\)

     

25 tháng 10 2019

Thay hộ mik cái dấu \(\le\)thành dấu \(\ge\)vs ak

5 tháng 10 2021

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

5 tháng 10 2021

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

 

25 tháng 10 2019

Bài 1:Tìm giá trị nhỏ nhất

A= x2+4x+100

A= (x\(^2\)+4x+4)+96

A= (x\(^2\)+2.x.2+2\(^2\))+96

A= (x+2)\(^2\)+96

Vì (x+2)\(^2\)0 x

(x+2)\(^2\)+96 ≥ 96 x

Vậy min A = 96 ⇔ x+2=0

⇔ x = -2

25 tháng 10 2019

B1 có bạn làm rồi

B2, B=-2.(x\(^2\)-3x+2)

=-2.(x\(^2\)-2.\(\frac{3}{2}\)x+\(\frac{9}{4}\)+2-\(\frac{9}{4}\))

=-2.[(x-\(\frac{3}{2}\))\(^2\)-\(\frac{1}{4}\)]

=-2.(x-\(\frac{3}{2}\))\(^2\)+\(\frac{1}{2}\)

Có -2.(x-\(\frac{3}{2}\))\(^2\)≤0∀x

⇒-2.(x-\(\frac{3}{2}\))\(^2\)+\(\frac{1}{2}\)\(\frac{1}{2}\)∀x

Dấu = xảy ra⇔x=\(\frac{3}{2}\)

GTLN của B=\(\frac{1}{2}\)