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DD
0
NT
1
TN
15 tháng 7 2016
a)x3+3x2+3x+1
=x3+3x2*1+3x*12+13
=(x+1)3
b)(x+y)2-9x2
=y2+2xy+x2-9x2
=y2-2xy+4xy-8x2
=y(y-2x)+4x(y-2x)
=(y-2x)(y+4x)
TA
25 tháng 7 2017
Bài 1 :
a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)
b) \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)
c) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
d) \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)
BÀi 2 :
a) \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)
\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)
b) \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)
c) \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)
\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)
\(=\left(b+c-a\right)\left(d-c^2\right)\)
BÀi 3 :
a) \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)
b) \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)
c) \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)
d) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
3 tháng 9 2017
a. \(x^3+3x^2-4=x^3+2x^2+x^2-4=x^2\left(x+2\right)+\left(x-2\right)\left(x+2\right)=\left(x+2\right)\left(x^2+x-2\right)\)
b. \(y^2+4y-12=y^2+4y+4-16=\left(y+2\right)^2-4^2=\left(y+2+4\right)\left(y+2-4\right)=\left(y+6\right)\left(y-2\right)\)
c. \(9x^2+6x-8=9x^2+6x+1-9=\left(3x+1\right)^2-3^2=\left(3x+1-3\right)\left(3x+1+3\right)=\left(3x-2\right)\left(3x+4\right)\)
d. \(2x^3-3x^2+3x-1=x^3-3x^2+3x-1+x^3=\left(x-1\right)^3+x^3=\left(x-1+x\right)\left[\left(x-1\right)^2-x\left(x-1\right)+x^2\right]=\left(2x-1\right)\left(x^2-2x+1-x^2+x+x^2\right)=\left(2x-1\right)\left(x^2-x+1\right)\)
e. \(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)=\left[a\left(x+y\right)+b\left(x+y\right)\right]\left[a\left(x-y\right)+b\left(y-x\right)\right]=\left(x+y\right)\left(a+b\right)\left(x-y\right)\left(a-b\right)=\left(x^2-y^2\right)\left(a^2-b^2\right)\)
31 tháng 7 2018
a) \(\left(9m^3-5p^2n\right)^2\)
b) \(\left(x^4-y^2\right)^3\)
c) \(\left(4x^5-3x^3\right)^3\)
7 tháng 8 2022
d: \(=\left(x+y\right)^3+3\left(x+y\right)^2+3\left(x+y\right)+1\)
\(=\left(x+y+1\right)^3\)
a: \(=\left(9m^3-5p^2n\right)^2\)
b: \(=\left(x^4-y^2\right)^3\)
c: \(=\left(4x^5-3x^3\right)^3\)
TN
25 tháng 7 2017
Bài 1:
\(a,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)
\(b,25-4x^2-4xy-y^2=25-\left(2x+y\right)^2\)
\(=\left(5-2x-y\right)\left(5+2x+y\right)\)
\(c,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\) \(d,x^2-4xy+4y^2-z^2+4tz-4t^2\)
\(=\left(x-2y\right)^2-\left(x-2t\right)^2=\left(x-2y-x+2t\right)\left(x-2y+x-2t\right)\)Bài 3,
\(a,x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(b,x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)\(c,x^3-5x^2+x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)
Ta có: \(x^2+1\ge1\Rightarrow x-5=0\Rightarrow x=5\)
\(d,x^4-2x^2+10x^3-20=0\)
\(\Leftrightarrow x^3\left(x-2\right)+x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)x\left(x^2+1\right)=0\)
ta có:
\(x^2+1\ge1\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
KT
20 tháng 7 2018
a) \(C=ax-ay-bx+by+\left(y-x\right)^2\)
\(=a\left(x-y\right)-b\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(a-b+x-y\right)\)
b) \(M=\left(5x-10\right)\left(x^2-1\right)\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=5\left(x-2\right)\left(x-1\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)^2\)
\(=\left(x-2\right)\left(x-1\right)\left[5\left(x+1\right)-3\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(5x+5-3x+3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(2x+8\right)\)
\(=2\left(x-2\right)\left(x-1\right)\left(x+4\right)\)