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Bài giải:
a) x2 – xy + x – y = (x2 – xy) + (x - y)
= x(x - y) + (x -y)
= (x - y)(x + 1)
b) xz + yz – 5(x + y) = z(x + y) - 5(x + y)
= (x + y)(z - 5)
c) 3x2 – 3xy – 5x + 5y = (3x2 – 3xy) - (5x - 5y)
= 3x(x - y) -5(x - y) = (x - y)(3x - 5).
\(a) x^2 - xy+x-y\) \(= (x^2 - xy) + ( x- y) \)
\(=x(x-y) + (x-y)\)
\(= (x-y) (x+1)\)
\(b) xz + yz - 5(x+y)\) \(= (xz + yz) - 5(x+y)\)
\(= z(x+y) - 5(x+y)\)
\(= (x+y) (z-5)\)
\(c) 3x^2 - 3xy - 5x +5y = (3x^2-3xy) - (5x-5y)\)
\(= 3x(x-y) - 5(x-y)\)
\(= (x-y)(3x-5)\)
1)
a) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
b) \(xz+yz-5\left(x+y\right)=z\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(z-5\right)\)
c) \(3x^2-3xy-5x+5y=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
Bài 2:
a) \(x^2+4x-y^2+4=\left(x^2+2\cdot x\cdot2+2^2\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
b) \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
a) \(5x-5y+ax-ay\)
\(\Leftrightarrow\) \(\left(5x+ax\right)-\left(5y+ay\right)\)
\(\Leftrightarrow\) \(x\left(5+a\right)-y\left(5+a\right)\)
\(\Leftrightarrow\) \(\left(5+a\right)\left(x-y\right)\)
a, Chứng minh \(x^3+y^3+z^3=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3\)
Biến đổi vế phải thì ta phải suy ra điều phải chứng minh
b, Ta có: \(a+b+c=0\)thì
\(a^3+b^3+c^3==\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
( Vì \(a+b+c=0\)nên \(a+b=-c\))
Theo giả thuyết \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
\(=xyz.\frac{3}{xyz}=3\)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
a) \(\frac{3m-6n}{10n-5m}\)
\(=\frac{-3\left(2n-m\right)}{5\left(2n-m\right)}=\frac{-3}{5}\)
b) \(\frac{y^3+y^2+4y+4}{y^2+2y-8}\)
\(=\frac{y^2\left(y+1\right)+4\left(y+1\right)}{y^2+2y+1-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y+1\right)^2-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y-2\right)\left(y+4\right)}\)
c) \(\frac{x^2-xy-xz+yz}{x^2+xy-xz-yz}\)
\(=\frac{x\left(x-y\right)-z\left(x-y\right)}{x\left(x+y\right)-z\left(x+y\right)}\)
\(=\frac{\left(x-z\right)\left(x-y\right)}{\left(x-z\right)\left(x+y\right)}\)
\(=\frac{x-y}{x+y}\)
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
a) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
b) \(x^2+y^2+2xy+yz+xz\)
\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
c) \(x^2-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)-1\)
\(=\left(x-5y\right)^2-1\)
\(=\left(x-5y-1\right)\left(x-5y+1\right)\)
d) \(ax^2-ax+bx^2-bx+a+b\)
\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)
\(=x^2(a+b)-x(a+b)+(a+b)\)
\(=(a+b)(x^2-x+1)\)
e)\(x^2-2y+3xz+x-2y+3z\)
\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)
\(=x(x+1)-2y(x-1)+3z(x+1)\)
\(=(x+1)(x-2y+3z)\)
f) \(xyz-xy-yz-xz+x+y+z-1\)
\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)
\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)
\(=(z-1)(xy-y-x+1)\)
\(=(z-1)[y(x-1)-(x-1)]\)
\(=(z-1)(x-1)(y-1)\)
_Học tốt_
bài này là phân tích đa thức thành nhân tử sao
a) x2 -xy+x-y
= ( x2-xy) +(x-y)
= x (x-y) +(x-y)
= (x-y) (x+1)
b) xz+yz-5( x+y)
= ( xz+yz)-5(x+y)
= z(x+y)-5(x+y)
= (x+y) (z-5)
c) 3x2-3xy-5x+5y
= ( 3x2 -3xy)-(5x+5y)
= 3x(x-y) - 5(x-y)
= (x-y) (3x-5)
uk