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Ta có: M=−x2−2x+5
=−(x2+2x−5)
=−(x2+2x+1)+6
=−(x+1)2+6
Vì −(x+1)2≤0∀x
⇒−(x+1)2+6≤6∀x
Dấu "=" xảy ra ⇔
Vậy
Đặt A=4x−x2+3
=−x2+4x+3=−(x2−4x−3)
=−(x2−4x+4−7)
=−[(x−2)2−7]
=−(x−2)2+7
Ta có: −(x−2)2≤0⇒−(x−2)2+7≤7
Dấu " = " khi (x−2)2=0⇔x=2
Vậy MAXA=7 khi x = 2
Ta có : P = x2 - 2x + 5 = x2 - 2x + 1 + 4 = (x - 1)2 + 4
Vì \(\left(x-1\right)^2\ge0\forall x\)
Suy ra : \(P=\left(x-1\right)^2+4\ge4\forall x\)
Nên : Pmin = 4 khi x = 1
b) Ta có Q = 2x2 - 6x = 2(x2 - 3x) = 2(x2 - 3x + \(\frac{9}{4}-\frac{9}{4}\) ) = \(2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)
Vì \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
SUy ra ; \(Q=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Vậy \(Q_{min}=-\frac{9}{2}\) khi \(x=\frac{3}{2}\)
Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(x^2-4x+1=x^2-2\cdot x\cdot2+4-4+1=\left(x-2\right)^2-4+1\)
\(=\left(x-2\right)^2-3\) \(\forall x\in Z\)
\(\Rightarrow A_{min}=-3khix=2\)
\(a,A=x^2-4x+1=x^2-2.2.x+2^2-3=\left(x-2\right)^2-3\ge-3\)
dấu = xảy ra khi x-2=0
=> x=2
Vậy MinA=-3 khi x=2
\(b,B=5-8x-x^2=-\left(x^2+8x+5\right)=-\left(x^2+2.4.x+4^2\right)+9=-\left(x+4\right)^2+9\le9\)
dấu = xảy ra khi x+4=0
=> x=-4
Vậy MaxB=9 khi x=-4
\(c,C=5x-x^2=-\left(x^2-5x\right)=-\left(x^2-\frac{2.x.5}{2}+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
dấu = xảy ra khi \(x-\frac{5}{2}=0\)
=> x=\(\frac{5}{2}\)
Vậy Max C=\(\frac{25}{4}\)khi x=\(\frac{5}{2}\)
\(E=\frac{1}{x^2+5x+14}=\frac{1}{x^2+\frac{2.x.5}{2}+\frac{25}{4}+\frac{31}{4}}=\frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{31}{4}}\)
\(\left(x+\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)
dấu = xảy ra khi \(x+\frac{5}{2}=0\)
=> x\(=-\frac{5}{2}\)
vì tử thức >0,mẫu thức nhỏ nhất và lớn hơn 0 => E lớnnhất khi mẫu thức nhỏ nhất
Vậy \(MaxE=\frac{31}{4}\)khi x\(=-\frac{5}{2}\)
a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:
\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)
b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)
\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
a. \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
vì \(\left(x-1\right)^2\ge0\) với mọi x
=> (x-1)^2 +4 \(\ge\) vợi mọi x
Pmin=4 <=> x-1=0 <=>x=1
1.
b)\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\) và \(y+3=0\)
\(\Leftrightarrow x=\frac{1}{2}\) và \(y=-3\)
Vậy GTNN của M là \(\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)và \(y=-3\)
a) ta có A = (2x-1)2+ ( x+2)= 4x2- 4x +1 +x+2= 4x2 -3x +3 = 4x2-2*2x* \(\frac{3}{4}\)+ \(\frac{9}{16}\)+ \(\frac{39}{16}\)
= (2x-\(\frac{3}{4}\))2+ \(\frac{39}{16}\)
=> (2x-\(\frac{3}{4}\))2>=0
=> A >= \(\frac{39}{16}\)
dấu = sảy ra khi x=\(\frac{3}{2}\)
vậy A(min) = \(\frac{39}{16}\) khi x=\(\frac{3}{2}\)
b) lm tương tự B(min)= -\(\frac{25}{4}\) khi x= \(\frac{5}{2}\)
c) đặt dấu trừ ra ngoài vậy C(max)=0 khi x=2
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)
..........
\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)
\(\Leftrightarrow\)\(x=-2040\)
Vậy phương trình có nghiệm là : x = -2040
giải câu b trc nha
= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009
vậy min=2009 khi x=1
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