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\(x^2-2x=24\)
<=> \(x^2-2x-24=0\)
<=> \( \left(x+4\right)\left(x-6\right)=0\)
<=> \(\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
Vậy....
\(a,\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow\left(x+2\right)^2+4-x^2=0\)
\(\Leftrightarrow\left(2+x\right)^2+\left(2-x\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left(2+x\right)\left(2+x+2-x\right)=0\)
\(\Leftrightarrow4\left(2+x\right)=0\)
\(\Leftrightarrow2+x=0\)
\(\Leftrightarrow x=-2\)
\(c,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)
\(\Leftrightarrow2x+255=0\)
\(\Leftrightarrow x=-127,5\)
\(a,x^2-2x+1=0\)
\(\left(x-1\right)^2=0\)
\(x-1=0\)
\(x=1\)
\(b,\left(x-3\right)\left(x+7\right)=0\)
\(\hept{\begin{cases}x-3=0\\x+7=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-7\end{cases}}}\)
\(c,x^4-4x^2=0\)
\(x^2\left(x^2-4\right)=0\)
\(x^2\left(x+2\right)\left(x-2\right)=0\)
\(\hept{\begin{cases}x^2=0\\x+2=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-2\\x=2\end{cases}}}\)
=.= hok tốt !!
a) x2-2x+1=0
(=) (x-1)2=0
(=) x=1
b (x-3)(x+7) =0
\(\left(=\right)\orbr{\begin{cases}x-3=0\\x+7=0\end{cases}}\left(=\right)\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)
c) (=) x2(x2-4) =0
\(\left(=\right)\orbr{\begin{cases}x^2=0\\x^2-4=0\end{cases}}\left(=\right)\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
1)
\(a;4-\left(a-b\right)^2=2^2-\left(a-b\right)^2=\left(2+a-b\right)\left(2-a+b\right)\)
\(b;\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)=5\left(x-y\right)\left(x+y\right)\)
\(c;16x^2-0,01=\left(4x\right)^2-0,1^2=\left(4x-0,1\right)\left(4x+0,1\right)\)
2)
\(x^2+16-8x=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
\(1.a)\)\(4-\left(a-b\right)^2=\left(2+a-b\right)\left(2-a+b\right)\)
\(b)\)\(\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(\left(5x-5y\right)\left(x+y\right)=5\left(x-y\right)\left(x+y\right)\)
\(c)\)\(16x^2-0,01=16x^2-\frac{1}{100}=\left(4x-\frac{1}{10}\right)\left(4x+\frac{1}{10}\right)\)
\(2.\)Ta có : \(x^2+16-8x=0=>\left(x-4\right)^2=0=>x-4=0=>x=4\)
Vậy \(x=4\)
i)
$I=x^4+4x^3-x^2-14x+6$
$=(x^4+4x^4+4x^2)-5x^2-14x+6$
$=(x^2+2x)^2-6(x^2+2x)+9+x^2-2x-3$
$=(x^2+2x-3)^2+(x^2-2x+1)-4$
$=(x-1)^2(x+3)^2+(x-1)^2-4$
$=(x-1)^2[(x+3)^2+1]-4\geq -4$
Vậy $I_{\min}=-4$ khi $(x-1)^2[(x+3)^2+1]=0\Leftrightarrow x=1$
k)
$K=x^4+2x^3-10x^2-16x+45$
$=(x^4+2x^3+x^2)-11x^2-16x+45$
$=(x^2+x)^2-12(x^2+x)+x^2-4x+45$
$=(x^2+x)^2-12(x^2+x)+36+(x^2-4x+4)+5$
$=(x^2+x-6)^2+(x-2)^2+5$
$=[(x-2)(x+3)]^2+(x-2)^2+5$
$=(x-2)^2[(x+3)^2+1]+5\geq 5$
Vậy $K_{\min}=5$ khi $(x-2)^2[(x+3)^2+1]=0\Leftrightarrow x=2$
g)
$G=x^4+4x^3+10x^2+12x+11$
$=(x^4+4x^3+4x^2)+6x^2+12x+11$
$=(x^2+2x)^2+6(x^2+2x)+11$
Đặt $x^2+2x=t$. Khi đó $t=x^2+2x=(x+1)^2-1\geq -1\Rightarrow t+1\geq 0$
$\Rightarrow G=t^2+6t+11=(t+1)^2+4(t+1)+7\geq 7$
Vậy $G_{\min}=7$ khi $t=-1\Leftrightarrow (x+1)^2=0\Leftrightarrow x=-1$
h)
$H=x^4-6x^3+x^2+24x+18$
$=(x^4-6x^3+9x^2)-8x^2+24x+18$
$=(x^2-3x)^2-8(x^2-3x)+18$
$=(x^2-3x)^2-8(x^2-3x)+16+2$
$=(x^2-3x-4)^2+2\geq 2$
Vậy $H_{\min}=2$ khi $x^2-3x-4=0\Leftrightarrow x=4$ hoặc $x=-1$
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
áp dụng công thức này mà lm câu a,b,e nhá:
\(A=ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}\\ \left[{}\begin{matrix}A\ge\dfrac{4ac-b^2}{4a}\left(với\text{ }\text{ }\text{ }a\ge0\right)\\A\le\dfrac{4ac-b^2}{4a}\left(với\text{ }a< 0\right)\end{matrix}\right.\)
\(C=x^2+2xy+y^2+4y^2=\left(x+y\right)^2+4y^2\ge0\)
đẳng thức xảy ra khi x=y=0
vậy MIN C=0 tại x=y=0
a: Ta có: \(x\left(2-x\right)+x^2+x=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)