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a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
Lời giải:
$\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:
$x=2k; y=3k$
Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.
$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$
a) \(\dfrac{4x-1}{3x^2y}-\dfrac{7x-1}{3x^2y}\)
\(=\dfrac{\left(4x-1\right)-\left(7x-1\right)}{3x^2y}\)
\(=\dfrac{4x-1-7x+1}{3x^2y}\)
\(=\dfrac{-3x}{3x^2y}\)
\(=\dfrac{-1}{xy}\)
b) \(\dfrac{4x+5}{2x-1}-\dfrac{5-9x}{2x-1}\)
\(=\dfrac{\left(4x+5\right)-\left(5-9x\right)}{2x-1}\)
\(=\dfrac{4x+5-5+9x}{2x-1}\)
\(=\dfrac{13x}{2x-1}\)
c) \(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)
\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)
\(=\dfrac{11x+\left(x-18\right)}{2x-3}\)
\(=\dfrac{11x+x-18}{2x-3}\)
\(=\dfrac{12x-18}{2x-3}\)
\(=\dfrac{6\left(2x-3\right)}{2x-3}\)
\(=\dfrac{6}{1}\)
\(=6\)
d) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)
\(=\dfrac{\left(2x-7\right)+\left(3x+5\right)}{10x-4}\)
\(=\dfrac{2x-7+3x+5}{10x-4}\)
\(=\dfrac{5x-2}{10x-4}\)
\(=\dfrac{5x-2}{2\left(5x-2\right)}\)
\(=\dfrac{1}{2}\)
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
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) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)
Suy ra: \(3x^2+7x-10=0\)
\(\Leftrightarrow3x^2-3x+10x-10=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)
a/ \(\dfrac{3x^2+7x-10}{x}=0\)
\(< =>3x^2+7x-10=0\)
\(< =>3x^2+10x-3x-10=0\)
\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)
\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)
\(< =>\left(3x+10\right)\left(x-1\right)=0\)
\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)
Vậy tập nghiệm của .....
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
