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N
3
17 tháng 9 2020
a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 15
= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 15
= ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 15 (*)
Đặt t = x2 + 5x + 4
(*) trở thành
t( t + 2 ) - 15
= t2 + 2t - 15
= t2 - 3t + 5t - 15
= t( t - 3 ) + 5( t - 3 )
= ( t - 3 )( t + 5 )
= ( x2 + 5x + 4 - 3 )( x2 + 5x + 4 + 5 )
= ( x2 + 5x + 1 )( x2 + 5x + 9 )
b) ( x + 2 )( x + 3 )2( x + 4 ) - 12
= [ ( x + 2 )( x + 4 ) ]( x + 3 )2 - 12
= ( x2 + 6x + 8 )( x2 + 6x + 9 ) - 12 (*)
Đặt t = x2 + 6x + 8
(*) trở thành
t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + 6x + 8 - 3 )( x2 + 6x + 8 + 4 )
= ( x2 + 6x + 5 )( x2 + 6x + 12 )
= ( x2 + x + 5x + 5 )( x2 + 6x + 12 )
= [ x( x + 1 ) + 5( x + 1 ) ]( x2 + 6x + 12 )
= ( x + 1 )( x + 5 )( x2 + 6x + 12 )
G
17 tháng 9 2020
a, Gọi\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt\(y=x^2+5x+4\)
\(\Rightarrow A=y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=\left(x-3\right)\left(x+5\right)\)
Hay\(A=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Vậy...
b,Gọi\(B=\left(x+2\right)\left(x+3\right)^2\left(x+4\right)-12\)
\(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)
Đặt\(z=x^2+6x+8\)
\(\Rightarrow B=z\left(z+1\right)-12\)
\(=z^2+z-12\)
\(=\left(z-3\right)\left(z+4\right)\)
Hay\(B=\left(x^2+6x+5\right)\left(x^2+6x+12\right)\)
Vậy...
Linz
SX
1
M
30 tháng 10 2015
\(4x^3-7x^2+3x\)
\(=4x^3-4x^2-3x^2+3x\)
\(=4x^2\left(x-1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(4x^2-3x\right)=x\left(x-1\right)\left(4x-3\right)\)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15\)
\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)-15\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-15\)
\(=\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)+1-16\)
\(=\left(x^2-5x+4+1\right)^2-4^2\)
\(=\left(x^2-4x+4+1-4\right)\left(x^2-4x+4+1+4\right)\)
\(=\left(x^2-4x+1\right)\left(x^2-4x+9\right)\)
M
1
HT
3
3 tháng 10 2019
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
3 tháng 10 2019
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
VN
3
7 tháng 2 2019
\(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(-5x+1\right)\)
M
0
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
15 tháng 11 2018
ta có (x-1)(x-2)(x-3)(x-4)-15=(x-1)(x-4)(x-2)(x-3)-15=\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15\)(*)
đặt \(t=x^2-5x+5\)thì pt (*) =\(\left(t-1\right)\left(t+1\right)-15=t^2-1-15\)\(=t^2-16=\left(t+4\right)\left(t-4\right)=\)\(\left(x^2-5x+5+4\right)\left(x^2-5x+5-4\right)=\)\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=t.\left(t+2\right)-15\)
\(=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2\)
\(=\left(t-3\right)\left(t+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Ta có :
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\)
\(=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]-15\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]-15\)(1)
Đặt \(x^2+5x+4=y\)thì (1) trở thành :
\(y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=y^2+5y-3y-15\)
\(=\left(y^2+5y\right)-\left(3y+15\right)\)
\(=y\left(y+5\right)-3\left(y+5\right)\)
\(=\left(y-3\right)\left(y+5\right)\)(2)
Thay \(y=x^2+5x+4\)thì (2) trở thành:
\(\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)