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3 tháng 10 2021

\(A=2\left(x^2+6x+36\right)-61=2\left(x+6\right)^2-61\ge-61\\ A_{min}=-61\Leftrightarrow x=-6\\ B=-\left(x^2-18x+81\right)+100=-\left(x-9\right)^2+100\le100\\ B_{max}=100\Leftrightarrow x=9\)

a: Ta có: \(A=2x^2+12x+11\)

\(=2\left(x^2+6x+\dfrac{11}{2}\right)\)

\(=2\left(x^2+6x+9-\dfrac{7}{2}\right)\)

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

Dấu '=' xảy ra khi x=-3

1 tháng 11 2018

kb nha

1 tháng 11 2018

a) \(B=-x^2+18x+19\)

\(B=-\left(x^2-2\cdot x\cdot9+9^2-100\right)\)

\(B=-\left[\left(x-9\right)^2-100\right]\)

\(B=100-\left(x-9\right)^2\le100\forall x\)( tự lí luận )

Dấu "=" xảy ra \(\Leftrightarrow x-9=0\Leftrightarrow x=9\)

Vậy Bmax = 100 khi và chỉ khi x = 9

b) \(A=2x^2+12x+11\)

\(A=2\left(x^2+6x+\frac{11}{2}\right)\)

\(A=2\left(x^2+2\cdot x\cdot3+3^2-\frac{7}{2}\right)\)

\(A=2\left[\left(x+3\right)^2-\frac{7}{2}\right]\)

\(A=2\left(x+3\right)^2-7\ge-7\forall x\)( tự lí luận )

Dấu "=" xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy Amin = -7 khi và chỉ khi x = -3

4 tháng 10 2021

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

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

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

b: Ta có: \(B=-2x^2+4x+1\)

\(=-2\left(x^2-2x-\dfrac{1}{2}\right)\)

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

\(=-2\left(x-1\right)^2+3\le3\forall x\)

Dấu '=' xảy ra khi x=1

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

1 tháng 9 2021

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)

b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)

c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

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

b: ta có: \(-x^2+5x+4\)

\(=-\left(x^2-5x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

NV
3 tháng 8 2021

\(M=-x^2+12x+8=-\left(x-6\right)^2+44\le44\)

\(M_{max}=44\) khi \(x=6\)

\(N=a^2+9b^2+5a-6b=\left(a+\dfrac{5}{2}\right)^2+\left(3b-1\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\)

\(N_{min}=-\dfrac{41}{4}\) khi \(\left(a;b\right)=\left(-\dfrac{5}{2};\dfrac{1}{3}\right)\)

\(Q=3\left(a-5\right)^2-82\ge-82\)

\(Q_{min}=-82\) khi \(a=5\)

19 tháng 12 2020

A= -x2+2x+3

=>A= -(x2-2x+3)

=>A= -(x2-2.x.1+1+3-1)

=>A=-[(x-1)2+2]

=>A= -(x+1)2-2

Vì -(x+1)≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)2=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

19 tháng 12 2020

B=x2-2x+4y2-4y+8

=> B= (x2-2x+1)+(4y2-4y+1)+6

=> B=(x-1)2+(2y+1)2+6

=> B lớn nhất=6 khi x=1 và y=-1/2