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b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)
b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)
c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)
\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)
d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)
hay \(N=y^2-x^2\)
Bài 1 :
a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)
TH1 : \(x^2-2x+3=0\)
\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm
TH2 : \(x-4=0\Leftrightarrow x=4\)
b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)
TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)
\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)
TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)
c, đưa về hệ đc ko ?
d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)
TH1 : \(x=0,74...\) ( bấm máy cx ra )
TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm
KL : vô nghiệm
Bài 2 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)
\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)
Vậy biểu thức ko phụ thuộc vào biến
b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-yx^3-y^2x^2-y^3x-y^4-x^4y^4\)
\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến
ChươngII *Dạng toán rútg gọn phân thức
Bài 1.Rút gọn phân thức
a. \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}=\dfrac{-3x\left(x-1\right)}{2\left(x-1\right)}=-\dfrac{3x}{2}\)
b.\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x.2xy^2}{4y^3.2xy^2}=\dfrac{3x}{4y^3}\)
c.\(\dfrac{23\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\dfrac{23\left(x-z\right)}{6}\)
Bài 2 rút gọn các phân thức sau:
a.\(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\dfrac{x+4}{x}\)(x khác 0,x khác 4)
b.\(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
( x \(\ne-3\) )
c.\(\dfrac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+y\right)}{y}\) (y+(x+y) khác 0)
d. \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)
(x khác y)
e.\(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
(x khác -y)
f.\(\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)(x khác y,y khác 0)
g.\(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
\ (b khác 0,x khác +-1)
h. \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4x}{5x^2}\)
(x khác 0,x khác y)
i.\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
(x+y+z khác 0)
k.\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3\right)^2+2x^3y^3+\left(y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
(x khác 0,x khác +-y)
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
cảm ơn bạn