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c) xét giá trị riêng
\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+xyz+xyz\)
\(=xy\left(x+y\right)+y^2z+yz^2+x^2z+xz^2+xyz+xyz\)
\(=xy\left(x+y\right)+y^2z+xyz+yz^2+xz^2+x^2z+xyz\)
\(=xy\left(x+y\right)+yz\left(x+y\right)+z^2\left(x+y\right)+xz\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+yz+z^2+xz\right)\)
\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
a) \(x^2-y^2-x-y\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
Bài 1:
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3-x+y\right)\)
\(=2\left(x-y\right)\left(2x+3+y\right)\)
Bài 2:
\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(3x-1-x-1\right)^2\)
\(=\left(2x-2\right)^2\)(1)
b) Thay \(x=\frac{9}{4}\)vào (1) ta được:
\(\left(2.\frac{9}{4}-2\right)^2\)
\(=\frac{25}{4}\)
Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)
Bài 3:
Ta có: \(M=x^2+4x+5\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)
Hay \(M\ge1;\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(M_{min}=1\Leftrightarrow x=-2\)
Bài 1 : trên là sai nha mình làm lại
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)
\(=2\left(x-y\right)\left(2x+4y\right)\)
\(=4\left(x-y\right)\left(x+2y\right)\)
Bài giải:
a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)
b) 2525x2 + 5x3 + x2y = x2 (2525 + 5x + y)
c) 14x2y – 21xy2 + 28x2y2 = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)
d) 2525x(y - 1) - 2525y(y - 1) = 2525(y - 1)(x - y)
e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]
= 10x(x - y) + 8y(x - y)
= 2(x - y)(5x + 4y)
a,\(3x-6y=3\left(x-2y\right)\)
b,\(x^2(\dfrac{2}{5}+5x+y)\)
c,\(7xy\left(2x-3y+4xy\right)\)
d,\(\dfrac{2}{5}x\left(y-1\right)-\dfrac{2}{5}y\left(y-1\right)\)
=\(\dfrac{2}{5}\left(y-1\right)\left(x-y\right)\)
e,\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)\)
\(2\left(x-y\right)\left(5x+4y\right)\)
\(=2\left[\left(x-3\right)^2-\dfrac{1}{16}\left(x-1\right)^2\right]\\ =2\left(x-3-\dfrac{1}{4}x+\dfrac{1}{4}\right)\left(x-3+\dfrac{1}{4}x-\dfrac{1}{4}\right)\\ =2\left(\dfrac{3}{4}x-\dfrac{11}{4}\right)\left(\dfrac{5}{4}x-\dfrac{13}{4}\right)\)