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a , \(-q^3+12q^2x-48qx^2+64x^3\)
\(=-\left(q^3-12q^2x+48qx^2-64x^3\right)\)
\(=\)\(-\left(q-4x\right)^3\)
b , x2 + 2xy - y2 - 9
= - ( x2 - 2xy + y2 ) - 9
= - ( x - y )2 - 9
= ( - x + y - 3 ) ( x - y + 3 )
3 , 1 - m2 + 2mn - n2
= 1 - ( m2 - 2mn + n2 )
= 1 - ( m - n )2
= ( 1 - m + n ) ( 1 + m - n )
4 , x3 - 8 + 6a2 - 12a
= x3 + 6a2 - 12a + 8
= x3 + 6a2 - 12a + 4 + 4
= x3 + ( 6a2 - 12a + 4 ) + 4
= x3 + ( 3a - 2 )2 + 4
= ( x + 3a - 2 + 2 ) ( x2 + 3a + 2 + 2 )
( Mai làm tiếp mấy ý sau '-' muộn rồi ~ )
5 , x2 - 2xy + y2 - xz - yz
= ( x2 - 2xy + y2 ) - ( xz + yz )
= ( x - y )2 - z ( x + y )
= ( x - y ) 2 - z ( x - y )
= ( x - y ) ( x - y - z )
6 , x2 - 4xy + 4y 2 - z2 + 4z - 4t2
=( x2 - 4xy + 4y 2 ) - (z2 - 4z +4 ) . t2
= ( x - y )2 - ( z - 2 )2 . t2
= ( x - y - z - 2 ) ( x - y + z - 2 ) t2
7 , 25 - 4x2 - 4xy - y2
= 25 + ( - 4x2 - 4xy + y2 )
= 25 + ( 2x - y )2
= ( 5 + 2x - y ) ( 5 + 2x + y )
8 ,
x3 + y3 + z3 - 3xyz
= (x+y)3 - 3xy (x - y ) + z3 - 3xyz
= [ ( x + y)3 + z3 ] - 3xy ( x + y + z )
= ( x + y + z )3 - 3z ( x + y )( x + y + z ) - 3xy ( x - y - z )
= ( x + y + z )[( x + y + z )2 - 3z ( x + y ) - 3xy ]
= ( x + y + z )( x2 + y2 + z2 + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= ( x + y + z)(x2 + y2 + z2 - xy - xz - yz)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
Bài 1 :
b, Ta có : \(4x^2-25-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
c, Ta có : \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=x\left(x+3\right)\left(x-2\right)\)
Bài 2 :
a, Để \(x^3+3x^2+3x-2⋮x+1\)
<=> \(x^3+1+3x^2+3x-3⋮x+1\)
<=> \(\left(x+1\right)^3-3⋮x+1\)
Ta thấy : \(\left(x+1\right)^3⋮x+1\)
<=> \(-3⋮x+1\)
<=> \(x+1\inƯ_{\left(3\right)}\)
<=> \(x+1=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{0,-2,2,-4\right\}\)
Vậy ...
b, Để \(2x^2+x-7⋮x-2\)
<=> \(2x^2-8x+8+9x-15⋮x-2\)
<=> \(2\left(x-2\right)^2+9x-15⋮x-2\)
Ta thấy : \(2\left(x-2\right)^2⋮x-2\)
<=> \(9x-15⋮x-2\)
<=> \(9x-18+3⋮x-2\)
Ta thấy : \(8\left(x-2\right)⋮x-2\)
<=> \(3⋮x-2\)
<=> \(x-2\inƯ_{\left(3\right)}\)
<=> \(x-2=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{3,1,5,-1\right\}\)
Vậy ...
Bài 3 :
a ) \(x\left(x-1\right)+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy...........
b ) \(3\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left(x-3\right)=0\Rightarrow x=3\)
Vậy............
Các câu sau tương tự
a) 2x2 + 4x + xy + 2y
= (2x2 + xy) + (4x + 2y)
= x(2x + y) + 2(2x + y)
= (x + 2)(2x + y)
b) x2 + xy - 7x - 7y
= x(x + y) - 7(x + y)
= (x - y)(x + y)
Phân tích đa thức thành nhân tử:
a) Ta có: \(3x^2-8xy+5y^2\)
\(=3x^2-3xy-5xy+5y^2\)
\(=3x\left(x-y\right)-5y\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5y\right)\)
b) Ta có: \(8xy^3+x\left(x-y\right)^3\)
\(=x\left[8y^3-\left(x-y\right)^3\right]\)
\(=x\left[2y-\left(x-y\right)\right]\left[4y^2+2y\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=x\left(2y-x+y\right)\left(4y^2+2xy-2y^2+x^2-2xy+y^2\right)\)
\(=x\left(3y-x\right)\left(3y^2+x^2\right)\)
c) Ta có: \(2x\left(x-3\right)-x+3\)
\(=2x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(2x-1\right)\)
d) Ta có: \(x^4-4x^3+4x^2\)
\(=x^2\left(x^2-4x+4\right)\)
\(=x^2\cdot\left(x-2\right)^2\)
e) Ta có: \(4x^2+4xy-4z^2+y^2-4z-1\)
\(=\left(4x^2+4xy+y^2\right)-\left(4z^2+4z+1\right)\)
\(=\left(2x+y\right)^2-\left(2z+1\right)^2\)
\(=\left(2x+y-2z-1\right)\left(2x+y+2z+1\right)\)
f) Ta có: \(x^2-2xy+y^2-x+y-6\)
\(=\left(x-y\right)^2-\left(x-y\right)-6\)
\(=\left(x-y\right)^2-3\left(x-y\right)+2\left(x-y\right)-6\)
\(=\left(x-y\right)\left(x-y-3\right)+2\left(x-y-3\right)\)
\(=\left(x-y-3\right)\left(x-y+2\right)\)
g) Ta có: \(x^2\left(x+3\right)^2-\left(x+3\right)^2-\left(x^2-1\right)\)
\(=x^2\left(x^2+6x+9\right)-\left(x^2+6x+9\right)-x^2+1\)
\(=\left(x^2-6x+9\right)\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-6x+9-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-6x+8\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-4\right)\)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
Dạng 1:
a) \(x^4+y^2-2x^2y=\left(x^2-y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a-b\right)\left(a+b\right)\)
c) \(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)^2\)
d) \(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ca-bc-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Dạng 2:
a) \(\left(7n-2\right)^2-\left(2n-7\right)^2\)
\(=\left(7n-2-2n+7\right)\left(7n-2+2n-7\right)\)
\(=\left(5n+5\right)\left(9n-9\right)\)
\(=45\cdot\left(n+1\right)\cdot\left(n-1\right)⋮3;5;9\) chứ không chia hết cho 7
Bạn xem lại đề.
b) \(n^3-n=n\left(n^2-1\right)=n\left(n-1\right)\left(n+1\right)\)
Vì \(n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên tích đó chia hết cho 2 và 3.
Mặt khác \(\left(2;3\right)=1\)
Do đó \(n\left(n-1\right)\left(n+1\right)⋮2.3=6\) ( đpcm