Tìm GTNN
A = x mũ 2 + 4x +100
Tìm GTLN
B = -2x mũ 2 + 6x - 4
~Cần gấp mn ơi >_<~
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NT
30 tháng 8 2017
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
19 tháng 9 2018
7(x - 3) - x(3 - x)
= (x - 3)(7 + x)
chỉ bt có v mà k bt có đúng k
SM
19 tháng 9 2018
1 ) 7 ( x - 3 ) - x ( 3 - x )
= 7 ( x - 3 ) + x ( x - 3 )
= ( x - 3 ) ( 7 + x )
2 ) 4x2 - 6x + 3 - 2x
= 4x2 - 2x - 6x + 3
= 2x ( 2x - 1 ) - 3 ( 2x - 1 )
= ( 2x - 1 ) ( 2x - 3 )
3 ) ( 4 - x ) - 4x + x2
= ( 4 - x ) - x ( 4 - x )
= ( 4 - x ) ( 1 - x )
4 ) x2 - 2xy + y2
= ( x - y )2
10 tháng 9 2020
\(A=x^2-6x+10=\left(x-3\right)^2+1\ge1\)
\(\Rightarrow A_{min}=1\Leftrightarrow x=3\)
\(B=4x^2-4x+25=\left(2x-1\right)^2+24\ge24\)
\(\Rightarrow B_{min}=24\Leftrightarrow x=\frac{1}{2}\)
\(C=3x^2+9x+12=3\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\ge\frac{21}{4}\)
\(\Rightarrow C_{min}=\frac{21}{4}\Leftrightarrow x=\frac{-3}{2}\)
21 tháng 9 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
21 tháng 9 2020
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
\(A=x^2+4x+100\)
\(A=x\left(x+4\right)+100\ge100\)
Dấu " = " xảy ra
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
Vậy Min A = 100 \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(B=-2x^2+6x-4\)
\(B=2x\left(3-x\right)-4\le-4\)
Dấu " = " xảy ra
\(\Leftrightarrow2x\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy Max B = -4 \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)