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Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!
a:6x-5-9x^2
=-(9x^2-6x+5)
=-(9x^2-6x+1+4)
=-(3x-1)^2-4<=-4
=>A>=2/-4=-1/2
Dấu = xảy ra khi x=1/3
b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)
2x^2-3x+2=2(x^2-3/2x+1)
=2(x^2-2*x*3/4+9/16+7/16)
=2(x-3/4)^2+7/8>=7/8
=>-1/2x^2-3x+2<=-1:7/8=-8/7
=>B<=-8/7+2=6/7
Dâu = xảy ra khi x=3/4
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
1.
Đặt \(x-2=t\ne0\Rightarrow x=t+2\)
\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)
\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)
2.
Đặt \(x-1=t\ne0\Rightarrow x=t+1\)
\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)
\(C_{max}=2\) khi \(t=3\) hay \(x=4\)
\(A=\dfrac{6x^2+21x+22}{x^2+4x+4}\)
\(=\dfrac{6\left(x^2+4x+4\right)-3x-2}{x^2+4x+4}\)
\(=6+\dfrac{-3x-2}{\left(x+2\right)^2}\)
\(=6+\dfrac{-3\left(x+2\right)+4}{\left(x+2\right)^2}\)
\(=6-\dfrac{3}{x+2}+\dfrac{4}{\left(x+2\right)^2}\)
-Đặt \(a=\dfrac{1}{x+2}\) thì:
\(A=6-3a+4a^2=\left(2a\right)^2-2.2a.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{87}{16}=\left(2a-\dfrac{3}{4}\right)^2+\dfrac{87}{16}\ge\dfrac{87}{16}\)
\(A_{min}=\dfrac{87}{16}\)\(\Leftrightarrow\left(2a-\dfrac{3}{4}\right)^2=0\Leftrightarrow2a-\dfrac{3}{4}=0\Leftrightarrow2a=\dfrac{3}{4}\)
\(\Leftrightarrow2.\dfrac{1}{x+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{x+2}=\dfrac{3}{8}\Leftrightarrow x+2=\dfrac{8}{3}\Leftrightarrow x=\dfrac{2}{3}\)
\(A=x^2+9x+25\)
\(=x^2+2x\frac{9}{2}+\frac{81}{4}+\frac{19}{4}\)
\(=\left(x+\frac{9}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
Dấu"="xảy ra khi \(\left(x+\frac{9}{2}\right)^2=0\Rightarrow x=\frac{-9}{2}\)
Vậy \(Min_A=\frac{19}{4}\Leftrightarrow x=\frac{-9}{2}\)
b,\(B=4x^2-8x+\frac{21}{2}\)
\(=4\left(x^2-2x+1\right)+\frac{13}{2}\)
\(=4\left(x-1\right)^2+\frac{13}{2}\ge\frac{13}{2}\forall x\)
Dấu"="xảy ra khi \(4\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Min_B=\frac{13}{2}\Leftrightarrow x=1\)
c,\(C=-x^2+2x+\frac{5}{2}\)
\(=-\left(x^2-2x-\frac{5}{2}\right)\)
\(=-\left(x^2-2x+1\right)+\frac{7}{2}\)
\(=-\left(x-1\right)^2+\frac{7}{2}\le\frac{7}{2}\forall x\)
Dấu"="xảy ra khi \(-\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy\(Max_C=\frac{7}{2}\Leftrightarrow x=1\)
Bài 1.
A = x2 + 9x + 25
= ( x2 + 9x + 81/4 ) + 19/4
= ( x + 9/2 )2 + 19/4 ≥ 19/4 ∀ x
Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2
=> MinA = 19/4 <=> x = -9/2
B = 4x2 - 8x + 21/2
= 4( x2 - 2x + 1 ) + 13/2
= 4( x - 1 )2 + 13/2 ≥ 13/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinB = 13/2 <=> x = 1
C = -x2 + 2x + 5/2
= -( x2 - 2x + 1 ) + 7/2
= -( x - 1 )2 + 7/2 ≤ 7/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MaxC = 7/2 <=> x = 1
D = -9x2 - 12x + 27/2
= -9( x2 + 4/3x + 4/9 ) + 35/2
= -9( x + 2/3 )2 + 35/2 ≤ 35/2 ∀ x
Đẳng thức xảy ra <=> x + 2/3 = 0 => x = -2/3
=> MaxD = 35/2 <=> x = -2/3
Bài 2.
a) 4x2 + 9y2 + 12x + 12y + 13 = 0
<=> ( 4x2 + 12x + 9 ) + ( 9y2 + 12y + 4 ) = 0
<=> ( 2x + 3 )2 + ( 3y + 2 )2 = 0 (*)
\(\hept{\begin{cases}\left(2x+3\right)^2\ge0\forall x\\\left(3y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(2x+3\right)^2+\left(3y+2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}2x+3=0\\3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{2}{3}\end{cases}}\)
=> x = -3/2 ; y = -2/3
b) 16x2 + 4y2 - 8x + 12y + 10 = 0
<=> ( 16x2 - 8x + 1 ) + ( 4y2 + 12y + 9 ) = 0
<=> ( 4x - 1 )2 + ( 2y + 3 )2 = 0 (*)
\(\hept{\begin{cases}\left(4x-1\right)^2\ge0\forall x\\\left(2y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(4x-1\right)^2+\left(2y+3\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}4x-1=0\\2y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{3}{2}\end{cases}}\)
=> x = 1/4 ; y = -3/2
ĐKXĐ: \(\dfrac{3}{2}\le x\le3\)
\(A=\sqrt{2x-3}+\sqrt{6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\)
\(A\ge\sqrt{2x-3+6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\ge\sqrt{3}\)
\(A_{min}=\sqrt{3}\) khi \(3-x=0\Rightarrow x=3\)
\(A=1.\sqrt{2x-3}+\sqrt{2}.\sqrt{6-2x}\le\sqrt{\left(1+2\right)\left(2x-3+6-2x\right)}=3\)
\(A_{max}=3\) khi \(2x-3=\dfrac{6-2x}{2}\Rightarrow x=2\)
\(E=2x^2+y^2-2xy-8x+24\)
\(=x^2+x^2+y^2-2xy-8x+16+8\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2-8x+16\right)+8\)
\(=\left(x-y\right)^2+\left(x-4\right)^2+8\)
Vậy \(E_{min}=8\Leftrightarrow x=y=4\)
Bài làm
E = 2x2 + y2 - 2xy - 8x + 24
E = ( x2 - 2xy + y2 ) + ( x2 - 8x + 16 ) + 8
E = ( x2 - 2xy + y2 ) + ( x2 - 2.4x + 42 ) + 8
E = ( x - y )2 + ( x - 4 )2 + 8 > 8
Dấu " = " xảy ra <=> E = 8
<=> x = 4; y = 4
Vậy E nhận giá trị nhỏ nhất là 8 khi x = 4 và y = 4
# Học tốt #
\(P=\frac{2}{-4x^2+8x-5}=\frac{2}{-\left(4x^2-8x+5\right)}\)
\(=\frac{2}{-\left(4x^2-8x+4+1\right)}\)\(=\frac{2}{-4\left(x+1\right)^2-1}\)
\(\ge\frac{2}{-1}=-2\)\(\Rightarrow P\ge-2\)
Dấu = khi \(x=-1\)
Vậy MinP=-2 khi x=-1
Cảm ơn bạn nhiều nha ! :)