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\(A=\dfrac{1}{-x^2+2x-2}\)
A min \(\Leftrightarrow\dfrac{1}{A}\)max
ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)
\(\dfrac{1}{A}\)max= -1 tại x=1
=> A min = -1 tại x=1
\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x2 nha bn)
B min \(\Leftrightarrow\dfrac{1}{B}\) max
ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)
\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2
\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2
\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)
A max \(\Leftrightarrow\dfrac{1}{A}\) min
\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)
\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)
\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)
B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)
B max \(\Leftrightarrow\dfrac{1}{B}\) min
\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)
\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)
=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)
Đây chỉ là gợi ý !! bn pải tự lí luận nha
tik 
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
\(E=\frac{5}{2x^2+3x+5}=\frac{5}{2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{35}{8}}=\frac{5}{2\left(x+\frac{3}{4}\right)^2+\frac{35}{8}}\le\frac{5}{\frac{35}{8}}=\frac{8}{7}\)
Nên GTLN của E là \(\frac{8}{7}\) đạt được khi x=\(-\frac{3}{4}\)
\(F=\frac{-2}{4x-x^2-5}=\frac{2}{x^2-4x+5}=\frac{2}{x^2-2.2x+4+1}=\frac{2}{\left(x-2\right)^2+1}\le\frac{2}{1}=2\)
Nên GTLN của F là 2 đạt được khi \(x=2\)
\(A=2x^2-7x+5=2\left(x^2-\dfrac{7}{2}x\right)+5=2\left(x^2-2.x.\dfrac{7}{4}+\dfrac{49}{16}\right)-\dfrac{9}{8}\\ =2\left(x-\dfrac{7}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
\(B=x^2-5x=x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\)
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
a) Ta có: \(2x^2+2x+3=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{5}{2}\)
\(=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)
\(\Rightarrow S\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)
Vậy \(S_{max}=\frac{6}{5}\Leftrightarrow\sqrt{2}x+\frac{1}{\sqrt{2}}=0\Leftrightarrow x=-\frac{1}{2}\)
b) Ta có: \(3x^2+4x+15=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\frac{2}{\sqrt{3}}+\frac{4}{3}+\frac{41}{3}\)
\(=\left(\sqrt{3}x+\frac{2}{\sqrt{3}}\right)^2+\frac{41}{3}\ge\frac{41}{3}\)
\(\Rightarrow T\le\frac{5}{\frac{41}{3}}=\frac{15}{41}\)
Vậy \(T_{max}=\frac{15}{41}\Leftrightarrow\sqrt{3}x+\frac{2}{\sqrt{3}}=0\Leftrightarrow x=\frac{-2}{3}\)
c) Ta có: \(-x^2+2x-2=-\left(x^2-2x+1\right)-1\)
\(=-\left(x-1\right)^2-1\le-1\)
\(\Rightarrow V\ge\frac{1}{-1}=-1\)
Vậy \(V_{min}=-1\Leftrightarrow x-1=0\Leftrightarrow x=1\)
d) Ta có: \(-4x^2+8x-5=-\left(4x^2-8x+5\right)\)
\(=-\left(4x^2-8x+4\right)-1\)
\(=-\left(2x-2\right)^2-1\le-1\)
\(\Rightarrow X\ge\frac{2}{-1}=-2\)
Vậy \(X_{min}=-2\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
1. D = 3( x2 - 2x.1/3 + 1/9) -1/3 +1
GTNN D = 5/6
dài quá, nản quá
\(A=x^2-8x+1=\left(x^2-8x+16\right)-15=\left(x+4\right)^2-15\)
Ta có \(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2-15\le-15\)
\(\Rightarrow Max_A=-15\Leftrightarrow\left(x+4\right)^2-15=-15\)
\(\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
a) ta có: A = x^2 - 8x + 1 = x^2 - 2.4.x + 16 - 15 = (x-4)^2 -15
=> giá trị nhỏ nhất của A = -15
b) ta có: B = 4 - x^2 + 4x = - (x^2 -4x + 4) + 8 = -(x-2)^2 +8
=> giá trị lớn nhất của B = 8
c) ta có: C = 3x^2 - 2x + 1
\(^2\ \)=> 3C =9 x^2 - 6x + 3
3C = 9x^2 - 2.3.x + 1 + 2
3C = (3x-1)^2 + 2
=> giá trị nhỏ nhất của 3C = 2 => giá trị nhỏ nhất của C = 2/3