Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
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)\(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu = khi \(x=\frac{-1}{2}\)
Vậy MinA=10 khi \(x=\frac{-1}{2}\)
b)\(B=3x^2-6x+1\)
\(=3x^2-6x+3-2\)
\(=3\left(x^2-2x+1\right)-2\)
\(=3\left(x-1\right)^2-2\ge-2\)
Dấu = khi \(x=1\)
Vậy MinB=-2 khi \(x=1\)
c)\(C=x^2-2x+y^2-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
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
a) \(A=2x^2+2x+3\)
\(A=2\left(x^2+x+\frac{3}{2}\right)\)
\(A=2\left[x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{5}{4}\right]\)
\(A=2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]\)
\(A=2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
b) Biến đổi mẫu thức :
\(3x^2+4x+15\)
\(=3\left(x^2+\frac{4}{3}x+5\right)\)
\(=3\left[x^2+2\cdot x\cdot\frac{2}{3}+\left(\frac{2}{3}\right)^2+\frac{41}{9}\right]\)
\(=3\left[\left(x+\frac{2}{3}\right)^2+\frac{41}{9}\right]\)
\(=3\left(x+\frac{2}{3}\right)^2+\frac{41}{3}\)
\(B=\frac{5}{3\left(x+\frac{2}{3}\right)^2+\frac{41}{3}}\ge\frac{5}{\frac{41}{3}}=\frac{15}{41}\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{2}{3}=0\Leftrightarrow x=\frac{-2}{3}\)
c) \(C=-x^2+2x-2\)
\(C=-\left(x^2-2x+2\right)\)
\(C=-\left(x^2-2\cdot x\cdot1+1^2+1\right)\)
\(C=-\left[\left(x-1\right)^2+1\right]\)
\(C=-1-\left(x-1\right)^2\le-1\)
Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
d) Biến đổi mẫu thức tương tự câu b)
\(P=\frac{xy}{\left|xy\right|}+\frac{x-y}{\left|x-y\right|}\cdot\left(\frac{x}{\left|x\right|}-\frac{y}{\left|y\right|}\right)\)
TH1: \(x,y>0\)
+) Xét \(x>y\): \(P=\frac{xy}{xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{x}-\frac{y}{y}\right)=1+1\cdot\left(1-1\right)=1\)
+) Xét \(x< y\): \(P=\frac{xy}{xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{x}-\frac{y}{y}\right)=1+\left(-1\right)\cdot\left(1-1\right)=1\)
TH2: \(x,y< 0\)
+) Xét \(x>y\): \(P=\frac{xy}{xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{-x}-\frac{y}{-y}\right)=1+1\cdot\left[-1-\left(-1\right)\right]=1\)
+) Xét \(x< y\): \(P=\frac{xy}{xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{-x}-\frac{y}{-y}\right)=1\)
TH3: \(x>0;y< 0\): \(P=\frac{xy}{-xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{x}-\frac{y}{-y}\right)=-1+1\cdot\left(1+1\right)=1\)
TH4: \(x< 0;y>0\): \(P=\frac{xy}{-xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{-x}-\frac{y}{y}\right)=-1+\left(-1\right)\cdot\left(-1-1\right)=1\)
Nói chung với mọi x, y thì P = 1
+) \(A=x^2+2x-9=x^2+2x+1-10=\left(x+1\right)^2-10\ge-10\)
Min A = -10 \(\Leftrightarrow x=-1\)
+) \(B=x^2+5x-1=x^2+5x+\frac{25}{4}-\frac{29}{4}=\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\ge\frac{-29}{4}\)
Min B = -29/4 \(\Leftrightarrow x=\frac{-5}{2}\)
+) \(C=x^2+4x=x^2+4x+4-4=\left(x+2\right)^2-4\ge-4\)
Min C = -4 \(\Leftrightarrow x=-2\)
+) \(D=x^2-8x+17=x^2-8x+16+1=\left(x-4\right)^2+1\ge1\)
Min D = 1 \(\Leftrightarrow x=4\)
+) \(E=x^2-7x+1=x^2-7x+\frac{49}{4}-\frac{45}{4}=\left(x-\frac{7}{2}\right)-\frac{45}{4}\ge-\frac{45}{4}\)
Min E = -45/4 \(\Leftrightarrow x=\frac{7}{2}\)
A = x2 + 2x - 9
= ( x2 + 2x + 1 ) - 10
= ( x + 1 )2 - 10 ≥ -10 ∀ x
Đẳng thức xảy ra <=> x + 1 = 0 => x = -1
=> MinA = -10 <=> x = -1
B = x2 + 5x - 1
= ( x2 + 5x + 25/4 ) - 29/4
= ( x + 5/2 )2 - 29/4 ≥ -29/4 ∀ x
Đẳng thức xảy ra <=> x + 5/2 = 0 => x = -5/2
=> MinB = -29/4 <=> x = -5/2
C = x2 + 4x
= ( x2 + 4x + 4 ) - 4
= ( x + 2 )2 - 4 ≥ -4 ∀ x
Đẳng thức xảy ra <=> x + 2 = 0 => x = -2
=> MinC = -4 <=> x = -2
D = x2 - 8x + 17
= ( x2 - 8x + 16 ) + 1
= ( x - 4 )2 + 1 ≥ 1 ∀ x
Đẳng thức xảy ra <=> x - 4 = 0 => x = 4
=> MinD = 1 <=> x = 4
E = x2 - 7x + 1
= ( x2 - 7x + 49/4 ) - 45/4
= ( x - 7/2 )2 - 45/4 ≥ -45/4 ∀ x
Đẳng thức xảy ra <=> x - 7/2 = 0 => x = 7/2
=> MinE = -45/4 <=> x = 7/2
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
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) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
Câu 3:
\(B=-3\left(x^2-\dfrac{1}{3}x-\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{13}{36}\right)\)
\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}< =\dfrac{13}{12}\)
Dấu '=' xảy ra khi x=1/6
Bài 4:
\(C=\left(x+y\right)^2-4\left(x+y\right)+1\)
=3^2-4*3+1
=9+1-12
=-2
\(A=x^2-8x+1=\left(x^2-8x+16\right)-15=\left(x+4\right)^2-15\)
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