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1) D = x2 - 7x + 6
D = x2 - 2.\(\dfrac{7}{2}x+\dfrac{49}{4}+6-\dfrac{49}{4}\)
D = \(\left(x-\dfrac{7}{2}\right)^2\) - \(\dfrac{25}{4}\)
Do : \(\left(x-\dfrac{7}{2}\right)^2\)≥ 0 ∀x
⇒ \(\left(x-\dfrac{7}{2}\right)^2\) - \(\dfrac{25}{4}\) ≥ - \(\dfrac{25}{4}\)
⇒DMIN = - \(\dfrac{25}{4}\) ⇔ x = \(\dfrac{7}{2}\)
E = 4x2 - 7x + 8
E = 4( x2 - \(2.\dfrac{7}{8}x+\dfrac{49}{64}\)) - \(\dfrac{49}{16}\) + 8
E = 4( x - \(\dfrac{7}{8}\))2 + \(\dfrac{79}{16}\) ≥ \(\dfrac{79}{16}\)
⇒ EMIN = \(\dfrac{79}{16}\) ⇔ x = \(\dfrac{7}{8}\)
Phùng Khánh Linh
muốn gõ dấu \(\ge\) mà không dùng công cụ có hình phép lấy tổng thì làm như thế nào?
\(A=-x^2+2x+3=-\left(x^2-2x-3\right)\)
\(=-\left(x^2-2x+1-4\right)\)
\(=-\left[\left(x-1\right)^2-4\right]=-\left(x-1\right)^2+4\le4\)
Vậy \(A_{max}=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(B=-2x^2-4x=-2\left(x^2+2x\right)\)
\(=-2\left(x^2+2x+1-1\right)\)
\(=-2\left[\left(x+1\right)^2-1\right]=-\left(x+1\right)^2+2\le2\)
Vậy \(B_{max}=2\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(C=-x^2-6x+12=-\left(x^2+6x-12\right)\)
\(=-\left(x^2+6x+9-21\right)\)
\(=-\left[\left(x+3\right)^2-21\right]=-\left(x+3\right)^2+21\le21\)
Vậy \(C_{max}=21\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(D=-x^2+3x-1==-\left(x^2-3x+1\right)\)
\(=-\left(x^2-3x+\frac{9}{4}-\frac{5}{4}\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
Vậy \(D_{max}=\frac{5}{4}\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
A = 2.(x^2-4x+4) - 18 = 2.(x-2)^2 - 18 >= -18
Dấu "=" xảy ra <=> x-2 = 0 <=> x=2
Vậy Min A = -18 <=> x=2
2. Ta có: A = x2 - 6x + 5 = (x2 - 6x + 9) - 4 = (x - 3)2 - 4
Ta luôn có: (x - 3)2 \(\ge\)0 \(\forall\)x
=> (x - 3)2 - 4 \(\ge\)-4 \(\forall\)x
Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3
Vậy MinA = -4 tại x = 3
Ta có: B = 4x2 - 8x + 7 = 4(x2 - 2x + 1) + 3 = 4(x - 1)2 + 3
Ta luôn có: 4(x - 1)2 \(\ge\)0 \(\forall\)x
=> 4(x - 1)2 + 3 \(\ge\)3 \(\forall\)x
Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1
vậy MinB = 3 tại x = 1
Ta có: C = 2x2 + 4x - 6 = 2(x2 + 2x + 1) - 8 = 2(x + 1)2 - 8
Ta luôn có: 2(x + 1)2 \(\ge\)0 \(\forall\)x
=> 2(x + 1)2 - 8 \(\ge\)-8 \(\forall\)x
Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1
Vậy MinC = -8 tại x = -1
1/
\(A=x^2-6x+5\)
\(A=x^2-2\cdot3x+3^2-3^2+5\)
\(A=\left(x-3\right)^2-3^2+5\)
\(A=\left(x-3\right)^2-9+5\)
\(A=\left(x-3\right)^2-4\)
mà \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-4\ge-4\)
\(\Rightarrow GTNNA\left(x^2-6x+5\right)=-4\)
với \(\left(x-3\right)^2=0;x=3\)
\(B=4x^2-8x+7\)
\(B=4\left(x^2-2x+\frac{7}{4}\right)\)
\(B=4\left(x^2-2\cdot1x+1-1+\frac{7}{4}\right)\)
\(B=4\left(x-1\right)^2+3\)
\(\left(x-1\right)^2\ge0\Rightarrow4\left(x^2-1\right)^2+3\ge3\)
\(\Rightarrow GTNNB=3\)
với \(\left(x-1\right)^2=0;x=1\)
\(C=2x^2+4x-6\)
\(C=2\left(x^2+2x-3\right)\)
\(C=2\left(x^2+2\cdot1x+1-1-3\right)\)
\(C=\left(x+1\right)^2-8\)
có\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2-8\ge-8\)
\(\Rightarrow GTNNC=-8\)
với \(\left(x+1\right)^2=0;x=-1\)
2.
c) \(C=2x^2+4x-6=2\left(x^2+2x+1\right)-8\)
\(=2\left(x+1\right)^2-8\ge-8\forall x\)
Dấu"=" xảy ra<=> \(2\left(x+1\right)^2=0\Leftrightarrow x=-1\)
3.
c) \(C=-3x^2-6x+9=-3\left(x^2+2x+1\right)+12\)
\(=-3\left(x+1\right)^2+12\le12\forall x\)
Dấu "=" xảy ra<=> \(-3\left(x+1\right)^2=0\Leftrightarrow x=-1\)
\(2,GTNN\)
\(A=x^2-6x+5=x^2+6x+9-4\)
\(=\left(x+3\right)^2-4\ge-4\)
\(A_{min}=-4\Leftrightarrow\left(x+3\right)^2=0\Rightarrow x=-3\)
\(B=4x^2-8x+7=4\left(x^2-2x+\frac{7}{4}\right)\)
\(=4\left(x^2-2x+1+\frac{3}{4}\right)=4\left(x-1\right)^2+3\ge3\)
\(\Rightarrow B_{min}=3\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)
\(C=2x^2+4x-6=2\left(x^2+2x-3\right)\)
\(=2\left(x^2+2x+1-4\right)=2\left(x+1\right)^2-8\ge-8\)
\(\Rightarrow C_{min}=-8\Leftrightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)
\(3,GTLN\)
\(A=-x^2+2x-3=-\left(x^2-2x+3\right)\)
\(=-\left(x^2-2x+1-4\right)=-\left(x-1\right)^2+4\le4\)
\(A_{max}=4\Leftrightarrow-\left(x-1\right)^2=0\Rightarrow x=1\)
\(B=-9x^2+6x-4=-\left[9x^2-6x+4\right]\)
\(=-\left[\left(3x\right)^2-6x+1+3\right]=-\left(3x-1\right)^2-3\)
\(B_{max}=-3\Leftrightarrow-\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)
\(C=-3x^2-6x+9=-3\left(x^2+2x-3\right)\)
\(=-3\left(x^2+2x+1-4\right)=-3\left(x+1\right)^2+12\)
\(C_{max}=12\Leftrightarrow-3\left(x+1\right)^2=0\Rightarrow x=-1\)
Đề phải là tìm Min mới đúng nhé!
\(A=6x-x^2+1=-x^2+6x+1=\left(-x^2+6x+9\right)-8\)
Đặt \(K=\left(-x^2+6x+9\right)\) .Để A đạt GTNN thì K nhỏ nhất:
ta có: \(K=\left(-x^2+6x+9\right)=-3\left(-\frac{1}{3}x^2-2x-3\right)\ge0\) (1)
Từ (1) ta có: \(A=K-8\ge-8\)
Dấu "=" xảy ra khi \(\left(-\frac{1}{3}x^2-2x-3\right)=0\Leftrightarrow x=-3\)
Vậy \(A_{min}=-8\Leftrightarrow x=-3\)
Mấy bài kia làm tương tự
tthKo bt thì đg làm nhé
\(A=6x-x^2+1=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left[\left(x-3\right)^2\right]+10\le10\)
Vậy GTLN của A là 10\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)