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a: \(2x^3+x^2-13x+6\)
\(=2x^3-4x^2+5x^2-10x-3x+6\)
\(=\left(x-2\right)\left(2x^2+5x-3\right)\)
\(=\left(x-2\right)\left(2x^2+6x-x-3\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\)
b: \(2x^2+y^2-6x+2xy-2y+5=0\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)
=>x-2=0 và x+y-1=0
=>x=2 và y=-1
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
a, x2-5xy+2x-10y = (x2 + 2x)-(5xy+10y)
= x(x+2)-5y(x+2)
= (x+2)(x-5y)
b, x2-5x+4 = x2- x - 4x +4
= (x2-x)-(4x-4)
=x(x-1)-4(x-4)
=(x-1)(x-4)
\(a,x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(x+2\right)\)
\(b,x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=x\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
mk chỉnh lại đề nha:
\(x^2-x-6\)
\(=x^2-3x+2x-6\)
\(=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x+2\right)\)
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
a) \(x^2\left(x-3\right)+27-9x=0\)
\(x^2\left(x-3\right)+9\left(3-x\right)=0\)
\(x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\left(x^2-9\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=9\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=3\end{cases}}\Rightarrow x=3\)
vay \(x=3\)
a) \(x^2-4x+3\)
= \(x^2-3x-x+3\)
\(=\left(x^2-3x\right)-\left(x-3\right)\)
\(=x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
a)\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(x-1\right)\)
b)\(x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)
c)\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)
d)\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a. 2xy - x2 - y2 + 16
=(2xy-x2-y2)+16
=16-(x-y)2
=(4+x-y)(4-x+y)
b. x2 + x - 6
=x2+3x-2x-6
=x(x+3)-2(x+3)
=(x-2)(x+3)
c. x2 + 5x + 6
=x2+3x+2x+6
=x(x+3)+2(x+3)
=(x+2)(x+3)
Giải:
a) \(x^2+5x=6\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2+6x-x-6=0\)
\(\Leftrightarrow\left(x^2+6x\right)-\left(x+6\right)=0\)
\(\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)
Vậy ...
b) \(x^2-2015x+2014=0\)
\(\Leftrightarrow x^2-2014x-x+2014=0\)
\(\Leftrightarrow\left(x^2-2014x\right)-\left(x-2014\right)=0\)
\(\Leftrightarrow x\left(x-2014\right)-\left(x-2014\right)=0\)
\(\Leftrightarrow\left(x-2014\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2014=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2014\\x=1\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
\(a,x^2+5x=6\)
\(\Rightarrow x^2+5x-6=0\)
\(\Rightarrow x^2+6x-x-6=0\)
\(\Rightarrow\left(x^2-6x\right)-\left(x+6\right)=0\)
\(\Rightarrow x\left(x+6\right)-\left(x+6\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)
\(b,x^2+2015x+2014=0\)
\(\Rightarrow x^2+2015x+2015-1=0\)
\(\Rightarrow\left(x^2-1\right)+\left(2015x+2015\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)+2015\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-1+2015\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+2014\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+2014=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2014\end{matrix}\right.\)