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Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Câu b bài 1 :
B = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
Bài 2 :
a. A = 2 ( x3 + y3 ) - 3 ( x2 + y2 ) với x + y = 1
=> A = 2 ( x + y ) ( x2 - xy + y2 ) - 3 [ ( x + y )2 - 2xy ]
=> A = 2 [ ( x + y )2 - 3xy ] - 3 ( 1 - 2xy )
=> A = 2 ( 1 - 3xy ) - 3 + 6xy
=> A = 2 - 6xy - 3 + 6xy
=> A = - 1
B = x3 + y3 + 3xy với x + y = 1
=> B = ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
=> B = ( x + y )3 - 3xy ( x + y - 1 )
=> B = 13 - 3xy . 0
=> B = 1
Bài 1.
a) ( x - 1 )3 + ( 2 - x )( 4 + 2x + x2 ) + 3x( x + 2 ) = 16
<=> x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x = 16
<=> 9x + 7 = 16
<=> 9x = 9
<=> x = 1
b) ( x + 2 )( x2 - 2x + 4 ) - x( x2 - 2 ) = 15
<=> x3 + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 7
<=> x = 7/2
c) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 15
<=> ( x - 3 )[ ( x - 3 )2 - ( x2 + 3x + 9 ) + 9( x2 + 2x + 1 ) = 15
<=> ( x - 3 )( x2 - 6x + 9 - x2 - 3x - 9 ) + 9x2 + 18x + 9 = 15
<=> ( x - 3 ).(-9x) + 9x2 + 18x + 9 = 15
<=> -9x2 + 27x + 9x2 + 18x + 9 = 15
<=> 45x + 9 = 15
<=> 45x = 6
<=> x = 6/45 = 2/15
d) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 3
<=> x( x2 - 25 ) - ( x3 + 8 ) = 3
<=> x3 - 25x - x3 - 8 = 3
<=> -25x - 8 = 3
<=. -25x = 11
<=> x = -11/25
Bài 2.
a) A = 2( x3 + y3 ) - 3( x2 + y2 )
= 2( x + y )( x2 - xy + y2 ) - 3x2 - 3y2
= 2( x2 - xy + y2 ) - 3x2 - 3y2
= 2x2 - 2xy + 2y2 - 3x2 - 3y2
= -x2 - 2xy - y2
= -( x2 + 2xy + y2 )
= -( x + y )2
= -(1)2 = -1
b) B = x3 + y3 + 3xy
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 + 3xy
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
= ( x + y )3 - 3xy( x + y - 1 )
= 13 - 3xy( 1 - 1 )
= 1 - 3xy.0
= 1
Câu 1 : Tìm x :
1. \(A=x^2+4x-2\)
\(A=x^2+2.x.2+2^2-2^2-2\)
\(A=\left(x^2+4x+2^2\right)-4-2\)
\(A=\left(x+2\right)^2-6\)
\(\left(x+2\right)^2-6\ge-6\)
MIn A= -6 khi \(\left(x+2\right)^2=0\)
=> \(x+2=0hayx=-2\)
Vậy x=2
những câu tiếp theo làm tg tự như thế nhé
Câu 1:
a) Ta có: \(A=x^2+4x-2\)
\(=x^2+4x+4-6\)
\(=\left(x+2\right)^2-6\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: x=-2
b) Ta có: \(B=2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)
\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)
\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)
\(=2\left(x-1\right)^2+1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: x=1
c) Ta có: \(C=x^2+y^2-4x+2y+5\)
\(=x^2-4x+4+y^2+2y+1\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y+1\right)^2\ge0\forall y\)
Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy: x=2 và y=-1
Câu 2:
a) Ta có: \(A=-x^2+6x+5\)
\(=-\left(x^2-6x-5\right)\)
\(=-\left(x^2-6x+9-14\right)\)
\(=-\left[\left(x^2-6x+9\right)-14\right]\)
\(=-\left[\left(x-3\right)^2-14\right]\)
\(=-\left(x-3\right)^2+14\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(A=-x^2+6x+5\) là 14 khi x=3
b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)
\(=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)
\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)
Ta có: \(\left(3y-1\right)^2\ge0\forall y\)
\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)
Từ (1) và (2) suy ra
\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)
\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)
Câu 3:
a) Ta có: \(x^2+y^2-2x+4y+5=0\)
\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy: x=1 và y=-2
b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy: x=3 và y=-2