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6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
T
5 tháng 7 2019
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
5 tháng 7 2019
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
1 tháng 7 2016
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-12\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-12\)
Đăt \(a=x^2+5x+5\)
\(\Rightarrow x^2+5x+5=a-1\)(Trừ 1 cho 2 vế \(a=x^2+5x+5\))
\(\Rightarrow x^2+5x+6=a+1\)( Cộng 1 vào cả 2 vế \(a=x^2+5x+5\))
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-12=\left(a-1\right)\left(a+1\right)-12\)
\(=a^2-13\)
\(= \left(a-\sqrt{13}\right)\left(a+\sqrt{13}\right)\)
\(=\left(x^2+5x+5-\sqrt{13}\right)\left(x^2+5x+5+\sqrt{13}\right)\)
KQ
1
18 tháng 7 2018
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-12\)
\(=\left(x+2\right)\left(x+5\right)\left(x+4\right)\left(x+3\right)-12\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-12\)
Đặt \(x^2+7x+10=t\)
\(=t\left(t+2\right)-12\)
\(=t^2+2t-12\)
Làm tiếp nha
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
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
27 tháng 11 2020
a, Đặt x^2 + x = t
\(t^2+4t-12=\left(t-2\right)\left(t+6\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
b, Đặt x + 1 = t
\(t\left(t+1\right)\left(t+2\right)\left(t+3\right)-24=\left(t^2+t\right)\left(t^2+3t+2t+6\right)\)
\(\left(t^2+t\right)\left(t^2+5t+6\right)-24=t^4+5t^3+6t^2+t^3+5t^2+6t-24\)
\(=t^4+6t^3+11t^2+6t-24=\left(t^3+7t^2+18t+24\right)\left(t-1\right)\)
\(=\left(t-1\right)\left(t+4\right)\left(t^2+3t+6\right)=x\left(x+5\right)\left[\left(x+5\right)^2+3\left(x+5\right)+6\right]\)
Đặt x + 2 = t ta có biểu thứ mới :
\(t\left(t+1\right)^2\left(t+2\right)-12=t\left(t^2+2t+1\right)\left(t+2\right)-12\)
\(=\left(t^3+2t^2+t\right)\left(t+2\right)-12=t^4+2t^3+2t^3+4t^2+t^2+2t-12\)
\(=t^4+4t^3+5t^2+2t-12=\left(t-1\right)\left(t^3+5t^2+10t+12\right)\)
\(=\left(t-1\right)\left(t+3\right)\left(t^2+2t+4\right)=\left(x+1\right)\left(x+5\right)\left[\left(x+2\right)^2+2\left(x+2\right)+4\right]\)
( 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
= 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 + 5x + x + 5 )( x2 + 6x + 12 )
= [ x( x + 5 ) + ( x + 5 ) ]( x2 + 6x + 12 )
= ( x + 5 )( x + 1 )( x2 + 6x + 12 )