1. tìm giá trị Mã của
a, A= -x^2 + 4x-5
b) B= -2x^2 -6x+5
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\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)
a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" \(\Leftrightarrow x=-1\)
b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)
c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)
Dấu "=" \(\Leftrightarrow x=2\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x)
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1]
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4
Vay gia tri nho nhat P=4 khi x=1
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4]
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2
Vay gia tri nho nhat Q= -9/2 khi x= 3/2
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4
= ( x-1/2)^2 + (y+3)^2 +3/4
M>= 3/4
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7]
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7
Vay GTLN A=7 khi x=2
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4]
GTLN B= 1/4 khi x=1/2
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4)
= -2[(x-1/2)^2 +9/4]
GTLN N= -9/2 khi x=1/2
\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)
\(A=x^2-6x+10=x^2-2\cdot x\cdot3+3^2+1=\left(x-3\right)^2+1\ge1\)
Vậy GTNN của A bằng 1. Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
\(B=4x-x^2-5=-\left(x^2-2\cdot x\cdot2+2^2+1\right)=-\left(x-2\right)^2+1\le1\)
Vây GTLN của B bằng 1. Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(C=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
Vậy GTNN của C bằng 4. Dấu '=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(D=x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy GTNN của D bằng 3/4. Dấu '=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
\(A=-x^2+4x-5\\ A=-\left(x^2-4x+4\right)-1\\ A=-\left(x^2-2\cdot x\cdot2+2^2\right)-1\\ A=-\left(x-2\right)^2-1\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow-\left(x-2\right)^2\le0\forall x\\ \Leftrightarrow A=-\left(x-2\right)^2-1\le-1\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Max\right)}=-1\text{ }khi\text{ }x=2\)
\(B=-2x^2-6x+5\\ B=-2x^2-6x-\dfrac{9}{2}+\dfrac{19}{2}\\ B=-\left(2x^2+6x+\dfrac{9}{2}\right)+\dfrac{19}{2}\\ B=-2\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{2}\\ B=-2\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{19}{2}\\ B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\\ Do\text{ }\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow-2\left(x+\dfrac{3}{2}\right)^2\le0\forall x\\ \Leftrightarrow B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\le\dfrac{19}{2}\forall x\\ \text{Dấu }"="\text{ xảy }ra\text{ }khi:\\ \left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }B_{\left(Max\right)}=\dfrac{19}{2}\text{ }khi\text{ }x=-\dfrac{3}{2}\)
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Lời giải:
(A=-x^2+4x-5)
(A=-left(x^2-4x+5 ight))
(A=-left(x^2-4x+4+1 ight))
(A=-left(x^2-4x+4 ight)-1)
(A=-left(x-2 ight)^2-1le1)
Dấu "=" xảy ra khi: (x=2)
(B=-2x^2-6x+5)
(B=-2left(x^2+3x-dfrac{5}{2} ight))
(B=-2left(x^2+3x-dfrac{19}{4}+dfrac{9}{4} ight))
(B=-2left(x^2+3x+dfrac{9}{4} ight)+dfrac{19}{2})
(B=-2left(x+dfrac{3}{2} ight)^2+dfrac{19}{2}ledfrac{19}{2})
Dấu "=" xảy ra khi : (x=-dfrac{3}{2})