Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. x5 + x + 1
=x5-x2+x2+x+1
=(x5-x2)+(x2+x+1)
=x2(x3-1)+(x2+x+1)
=x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x-1)+1]
=(x2+x+1)(x3-x2+1)
2. x5 + x4 +1
=x5+x4+x3-x3+1
=(x5+x4+x3)-(x3-1)
=x3(x2+x+1)-(x-1)(x2+x+1)
=(x2+x+1)[x3-(x-1)]
=(x2+x+1)(x3-x+1)
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
a) \(4x^4-21x^2y^2+y^4=\left(4x^4+4x^2y^2+y^4\right)-25x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(5xy\right)^2=\left(2x^2-5xy+y^2\right)\left(2x^2+5xy+y^2\right)\)
b) \(x^5-5x^3+4x=x\left(x^4-5x^2+4\right)=x\left[\left(x^4-4x^2\right)-\left(x^2-4\right)\right]\)
\(=x\left[x^2\left(x^2-4\right)-\left(x^2-4\right)\right]=x\left(x^2-1\right)\left(x^2-4\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
c ) \(x^3+5x^2+3x-9=\left(x^3-x^2\right)+\left(6x^2-6x\right)+\left(9x-9\right)\)
\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x^2+6x+9\right)\left(x-1\right)=\left(x+3\right)^2\left(x-1\right)\)
d ) \(x^{16}+x^8-2=x^{16}-x^8+2x^8-2=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)
\(=\left(x^8+2\right)\left(x^8-1\right)=\left(x^8+2\right)\left(x^4-1\right)\left(x^4+1\right)\)
\(=\left(x^8+2\right)\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)=\left(x^8+2\right)\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\)
e ) \(x^3-11x^2+30x=0\)
\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)
\(\Leftrightarrow x\left[\left(x^2-5x\right)-\left(6x-30\right)\right]=0\)
\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)
\(\Leftrightarrow x\left(x-6\right)\left(x-5\right)=0\)
\(\Rightarrow x=0orx=5orx=6\) (or hoặc)
Vậy \(x\in\left\{0;5;6\right\}\)
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
a) x2( x - 1 ) - x + 1
= x2( x - 1 ) - ( x - 1 )
= ( x - 1 )( x2 - 1 )
= ( x - 1 )( x - 1 )( x + 1 )
= ( x - 1 )2( x + 1 )
b) ( a + b )3 - ( a - b )3
= ( a3 + 3a2b + 3ab2 + b3 ) - ( a3 - 3a2b + 3ab2 - b3 )
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3
= 6a2b + 2b3
= 2b( 3a2 + b )
c) 6x( x - 3 ) + 9 - 3x2
= 6x2 - 18x + 9 - 3x2
= 3x2 - 18x + 9
= 3( x2 - 6x + 3 )
d) x( x - y ) - 5x + 5y
= x( x - y ) - ( 5x - 5y )
= x( x - y ) - 5( x - y )
= ( x - y )( x - 5 )
e) 3( x + 4 ) - x2 - 4x
= 3( x + 4 ) - ( x2 + 4x )
= 3( x + 4 ) - x( x + 4 )
= ( x + 4 )( 3 - x )
f) x2 + 4x - y2 + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
g) x2 + 5x
= x( x + 5 )
h) -x2 + 2x + 2y + y2
= ( y2 - x2 ) + ( 2x + 2y )
= ( y - x )( y + x ) + 2( x + y )
= ( x + y )( y - x + 2 )
Bài 2:
a, \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy...
b, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
c, \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
Bài 3:
1, Đặt \(A=x^2+\dfrac{1}{2}x+\dfrac{1}{16}=x^2+\dfrac{1}{4}.x.2+\dfrac{1}{16}\)
\(=\left(x+0,25\right)^2\)
Thay x = 49,75 vào A ta có:
\(A=50^2=2500\)
2, tương tự
bài 1 bn ơi