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b) \(\hept{\begin{cases}\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}\\\frac{3}{x^2-9}=\frac{3}{\left(x+3\right)\left(x-3\right)}\end{cases}}\)
\(\Rightarrow MTC=2\left(x+3\right)\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{5}{2\left(x+3\right)}=\frac{5\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}\\\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{6}{2\left(x-2\right)\left(x+3\right)}\end{cases}}\)
CÒn lại tương tự nhé !
a. \(x^2y^3.35xy=5.7x^3y^4\)
\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)
\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)
\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)
\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)
\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)
\(\Rightarrowđpcm\)
\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)
\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)
a) Tìm MTC:
2x + 6 = 2(x + 3)
x2 – 9 = (x – 3)(x + 3)
MTC = 2(x – 3)(x + 3) = 2(x2 – 9)
Nhân tử phụ:
2(x – 3)(x + 3) : 2(x + 3) = x – 3
2(x – 3)(x + 3) : (x2 – 9) = 2
Qui đồng:
b) Tìm MTC:
x2 – 8x + 16 = (x – 4)2
3x2 – 12x = 3x(x – 4)
MTC = 3x(x – 4)2
Nhân tử phụ:
3x(x – 4)2 : (x – 4)2 = 3x
3x(x – 4)2 : 3x(x – 4) = x – 4
Qui đồng:
\(\dfrac{3x}{x+3}=\dfrac{3x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x^2-9x}{x^2-9}\)
\(\dfrac{x-1}{x-3}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2x-3}{x^2-9}\)
\(x^2+9=\dfrac{\left(x^2+9\right)\left(x^2-9\right)}{x^2-9}=\dfrac{x^4-81}{x^2-9}\)
