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Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)
b)\(=\frac{3x\left(x+y\right)}{y}\)
c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)
b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)
c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)
h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)
j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)
Câu b) bạn xem lại nhé.
Học tốt ^3^
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
ko ghi đề bài nha làm luôn
a) \(\frac{\left(2x+2y\right)+\left(5x+5y\right)}{\left(2x+2y\right)-\left(5x+5y\right)}=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\frac{\left(2+5\right)\left(x+y\right)}{\left(2-5\right)\left(x+y\right)}=\frac{-7}{3}\)
b)\(\frac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\frac{4x}{5x^2}=\frac{4}{5x}\)
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha