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Bài giải:
a) x2 – 3x + 2 = a) x2 – x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)
Hoặc x2 – 3x + 2 = x2 – 3x - 4 + 6
= x2 - 4 - 3x + 6
= (x - 2)(x + 2) - 3(x -2)
= (x - 2)(x + 2 - 3) = (x - 2)(x - 1)
b) x2 + x – 6 = x2 + 3x - 2x – 6
= x(x + 3) - 2(x + 3)
= (x + 3)(x - 2).
c) x2 + 5x + 6 = x2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
\(x^4-5x^3+7x^2-6\)
\(=x^4-3x^3+3x^2-2x^3+6x^2-6x-2x^2+6x-6\)
\(=x^2\left(x^2-3x+3\right)-2x\left(x^2-3x+3\right)-2\left(x^2-3x+3\right)\)
\(=\left(x^2-3x+3\right)\left(x^2-2x-2\right)\)
\(\left(x^2-x+6\right)^2+\left(x-3\right)^2\)
\(=x^4+x^2+36-2x^3-12x+12x^2+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4-2x^3+5x^2+9x^2-18x+45\)
\(=x^2\left(x^2-2x+5\right)+9\left(x^2-2x+5\right)=\left(x^2-2x+5\right)\left(x^2+9\right)\)
Bài này hay và khó đấy. Chúc bạn học tốt.
a) x2 – 3x + 2 = a) x2 – x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)
Hoặc x2 – 3x + 2 = x2 – 3x - 4 + 6
= x2 - 4 - 3x + 6
= (x - 2)(x + 2) - 3(x -2)
= (x - 2)(x + 2 - 3) = (x - 2)(x - 1)
b) x2 + x – 6 = x2 + 3x - 2x – 6
= x(x + 3) - 2(x + 3)
= (x + 3)(x - 2).
Câu 1:
a) \(2x^2+5x-3=\left(2x^2+6x\right)-\left(x+3\right)\)
\(=2x\left(x+3\right)-\left(x+3\right)=\left(x+3\right)\left(2x-1\right)\)
b) \(x^4+2009x^2+2008x+2009\)
\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)
c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5-\sqrt{5}\\x=-5+\sqrt{5}\end{cases}}\)
Câu 1.
a) 2x2 + 5x - 3 = 2x2 + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )
b) x4 + 2009x2 + 2008x + 2009
= x4 + 2009x2 + 2009x - x + 2009
= ( x4 - x ) + ( 2009x2 + 2009x + 2009 )
= x( x3 - 1 ) + 2009( x2 + x + 1 )
= x( x - 1 )( x2 + x + 1 ) + 2009( x2 + x + 1 )
= ( x2 + x + 1 )[ x( x - 1 ) + 2009 ]
= ( x2 + x + 1 )( x2 - x + 2009 )
c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )
Câu 2.
3x2 + x - 6 - √2 = 0
<=> ( 3x2 - 6 ) + ( x - √2 ) = 0
<=> 3( x2 - 2 ) + ( x - √2 ) = 0
<=> 3( x - √2 )( x + √2 ) + ( x - √2 ) = 0
<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0
<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)
+) x - √2 = 0 => x = √2
+) 3( x + √2 ) + 1 = 0
<=> 3( x + √2 ) = -1
<=> x + √2 = -1/3
<=> x = -1/3 - √2
Vậy S = { √2 ; -1/3 - √2 }
Câu 3.
A = x( x + 1 )( x2 + x - 4 )
= ( x2 + x )( x2 + x - 4 )
Đặt t = x2 + x
A = t( t - 4 ) = t2 - 4t = ( t2 - 4t + 4 ) - 4 = ( t - 2 )2 - 4 ≥ -4 ∀ t
Dấu "=" xảy ra khi t = 2
=> x2 + x = 2
=> x2 + x - 2 = 0
=> x2 - x + 2x - 2 = 0
=> x( x - 1 ) + 2( x - 1 ) = 0
=> ( x - 1 )( x + 2 ) = 0
=> x = 1 hoặc x = -2
=> MinA = -4 <=> x = 1 hoặc x = -2
trước tiên mik xin l các bn vì mik vt sai đề:5x4-x2-6
5x4-x2-6
=5x4+5x2-(6x2+6)
=5x2(x2+1)-6(x2+1)
=(5x2-6)(x2+1)
ai ko hiểu thì ? đừng k sai nha!
câu a:
\(=x^2+6x-x+6\)
\(=\left(x^2-x\right)-\left(6x-6\right)\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
câu b:
\(=x^2+5x-x-5\)
\(=x^2-x+5x-5\)
\(=x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x+5\right)\left(x-1\right)\)
a, x2 + 5x +6
= x2 - 6x-x +6
= x(x-6)-(x-6)
=( x-1)(x-6)
b, x2+4x-5
= x2+ 5x -x -5
= x(x+5)-(x+5)
=(x-1)(x+5)
a) \(2x^2-5x-12\)
\(=2x^2-8x+3x-12\)
\(=2x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+3x^2+2x\right)+\left(2x^2+6x+4\right)\)
\(=x\left(x^2+3x+2\right)+2\left(x^2+3x+2\right)\)
\(=\left(x^2+3x+2\right)\left(x+2\right)\)
\(=\left(x^2+x+2x+2\right)\left(x+2\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
c) \(x^4+x^2+1\)
\(=\left(x^4-x^3+x^2\right)+\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\)
\(=x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^2-5x+6=0\)
=>\(x^2-2x-3x+6=0\)
=>x(x-2)-3(x-2)=0
=>(x-2)(x-3)=0
=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)