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a) \(x^2\)\(+3x+7\)
=\(x^2\)\(+2.x.\frac{3}{2}\)\(+\frac{9}{4}\)\(+\frac{19}{4}\)
=\(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)
Vì \(\left(x+\frac{3}{2}\right)^2\)\(\ge0\)
Nên \(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)\(\ge\frac{19}{4}\)
Dấu "=" xảy ra khi:
\(x+\frac{3}{2}\)\(=0\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy GTNN của \(x^2\)\(+3x+7\) là \(\frac{19}{4}\) khi \(x=-\frac{3}{2}\)
b) \(-9x^2+12x-15\)
=\(-\left(9x^2-12x+15\right)\)
=\(-\left(\left(3x\right)^2-2.3x.2+4+11\right)\)
=\(-\left(\left(3x-2\right)^2+11\right)\)
=\(-\left(3x-2\right)^2-11\)
Vì \(\left(3x-2\right)^2\)\(\ge0\)
Nên \(-\left(3x-2\right)^2-11\le-11\)
Dấu "=" xảy ra khi:
\(3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy GTLN của \(-9x^2+12x-15\) là \(-11\) khì \(x=\frac{2}{3}\)
c) \(11-10x-x^2\)
=\(-\left(x^2+10x-11\right)\)
=\(-\left(x^2+2.x.5+25-36\right)\)
=\(-\left(\left(x+5\right)^2-36\right)\)
=\(-\left(x+5\right)^2+36\)
Vì \(\left(x+5\right)^2\ge0\)
Nên \(-\left(x+5\right)^2+36\le36\)
Dấu "=" xảy ra khi:
\(x+5=0\)
\(\Rightarrow x=-5\)
Vậy GTLN \(11-10x-x^2\) là \(36\) khi \(x=-5\)
d)\(x^4+x^2+2\)
=\(\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)
=\(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\)
Vì \(\left(x^2+\frac{1}{2}\right)^2\ge0\)
Nên \(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
Dấu "=" xảy ra khi:
\(x^2+\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{\sqrt{2}}\)
Vậy GTNN của \(x^4+x^2+2\) là \(\frac{7}{4}\) khi \(x=\frac{1}{\sqrt{2}}\)
a) \(x^2+3x+7=x^2+2.1,5x+1,5^2+4,75=\left(x+1,5\right)^2+4,75\ge4,75\)
Đẳng thức xảy ra khi : \(x+1,5=0\Rightarrow x=-1,5\)
Vậy giá trị nhỏ nhất của x2 + 3x + 7 là 4,75 khi x = -1,5
b) \(-9x^2+12x-15=-\left(9x^2-12x+15\right)=-\left[\left(3x\right)^2-2.2.3x+2^2+11\right]\)
\(=-\left[\left(3x-2\right)^2+11\right]=-\left(3x-2\right)^2-11\le-11\)
Đẳng thức xảy ra khi : \(3x-2=0\Rightarrow x=\frac{2}{3}\)
Vậy giá trị lớn nhất của -9x2 +12x - 15 là -11 khi \(x=\frac{2}{3}\)
c) \(11-10x-x^2=-x^2-10x+11=-\left(x^2+10x-11\right)=-\left(x^2+2.5x+5^2-36\right)\)
\(=-\left[\left(x+5\right)^2-36\right]=-\left(x+5\right)^2+36\le36\)
Đẳng thức xảy ra khi : \(x+5=0\Rightarrow x=-5\)
Vậy giá trị lớn nhất của 11 - 10x -x2 là 36 khi x = -5.
\(E=-16x^2+3x-3=-\left(4x-\frac{3}{8}\right)^2-\frac{183}{64}\le\frac{-183}{64}\)
Vậy \(MaxE=\frac{-183}{64}\) khi \(x=\frac{3}{32}\)
Bạn xem lại đề phần \(F\) nhé.
\(G=-3x^2-9x+2=-3\left(x^2+3x-\frac{2}{3}\right)=-3[x^2+2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2]+\frac{35}{4}\)
\(=-3\left(x+\frac{3}{2}\right)^2+\frac{35}{4}\le\frac{35}{4}\forall x\)
Vậy \(MaxG=\frac{35}{4}\) khi: \(\left(x+\frac{3}{2}\right)^2=0\Rightarrow x=\frac{-3}{2}\)
\(H=-7x^2+14x-3=-7\left(x^2-2x+\frac{3}{7}\right)=-7\left(x^2-2x+1\right)+4=-7\left(x-1\right)^2+4\le4\forall x\)
Vậy \(MaxH=4\) khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)
a: Ta có: \(B=x^2-4x+6\)
\(=x^2-4x+4+2\)
\(=\left(x-2\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=2
a,\(x^2-6x-17=x^2-2\cdot3x+9-26=\left(x-3\right)^2-26\ge-26\)
b, \(x^2-10x=x^2-2\cdot5x+25-25=\left(x-5\right)^2-25\ge-25\)
c,\(3x^2-12x+5=3x^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+12-7=\left(\sqrt{3}x-2\sqrt{3}\right)^2-7\ge-7\)
d,\(2x^2-x-1=2x^2-2\cdot\sqrt{2}x\cdot\dfrac{1}{2\sqrt{2}}+\dfrac{1}{8}-\dfrac{9}{8}=\left(\sqrt{2}x-\dfrac{1}{2\sqrt{2}}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
e,\(x^2+y^2-8x+4y+27=x^2-2\cdot4x+16+y^2+2\cdot2y+4+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\)
f,\(x\left(x-6\right)=x^2-6x=x^2-2\cdot3x+9-9=\left(x-3\right)^2-9\ge-9\)
h,\(\left(x-2\right)\cdot\left(x-5\right)\cdot\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)=\left(x^2-7x\right)^2-100\ge-100\)
Mình giúp tính biểu thức thôi
còn lại bạn tự làm nhé
c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)
\(\Leftrightarrow V\ge-1\forall x\)
Dấu '=' xảy ra khi x=1
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Tìm GTNN
A = x2 - 10x + 3 = ( x2 - 10x + 25 ) - 22 = ( x - 5 )2 - 22 ≥ -22 ∀ x
Dấu "=" xảy ra khi x = 5
=> MinA = -22 <=> x = 5
B = 3x2 + 7x - 2 = 3( x2 + 7/3x + 49/36 ) - 73/12 = 3( x + 7/6 )2 - 73/12 ≥ -73/12 ∀ x
Dấu "=" xảy ra khi x = -7/6
=> MinB = -73/12 <=> x = -7/6
Tìm GTLN
A = -9x2 + 12x - 5 = -9( x2 - 4/3x + 4/9 ) - 1 = -9( x - 2/3 )2 - 1 ≤ -1 ∀ x
Dấu "=" xảy ra khi x = 2/3
=> MaxA = -1 <=> x = 2/3
B = -2x2 - 3x + 7 = -2( x2 + 3/2x + 9/16 ) + 65/8 = -2( x + 3/4 )2 + 65/8 ≤ 65/8 ∀ x
Dấu "=" xảy ra khi x = -3/4
=> MaxB = 65/8 <=> x = -3/4