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a) \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
b) \(125x^6-27x^9=\left(5x^2-3x^3\right)\left(25x^4+15x^5+9x^6\right)\)
c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
a: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
b: \(27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: \(y^6+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
d: \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(\frac{-x^6}{125}-\frac{y^3}{64}\)
\(=\frac{-\left(x^2\right)^3}{5^3}-\frac{y^3}{4^3}\)
\(=\left(\frac{-x^2}{5}\right)^3-\left(\frac{y}{4}\right)^3\)
\(=\left(\frac{-x^2}{5}-\frac{y}{4}\right)\cdot\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)
Tham khảo nhé~
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
a, \(x^3+8=x^3+2x^2-2x^2-4x+4x+8\)
\(=x^2.\left(x+2\right)-2x.\left(x+2\right)+4.\left(x+2\right)\)
\(=\left(x+2\right).\left(x^2-2x+4\right)\)
D) 64x^3-1/8y^3
= (4x)^3 + (1/2y)^3
= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]
E) 125x^6-27y^9
( câu này mik chưa rõ nên vx chưa tek giải cho bn )
HOk tốt nhé
a/ \(\frac{7x-14y}{x^2-4y^2}=\frac{7\left(x-2y\right)}{x^2-\left(2y\right)^2}=\frac{7\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{7}{x+2y}.\)
b/ \(\frac{1-\frac{2y}{x}+\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x^2-2xy+y^2}{x^2}}{\frac{y-x}{xy}}=\frac{\left(x-y\right)^2}{x^2}.\frac{xy}{-\left(x-y\right)}=-\frac{y\left(x-y\right)}{x}=\frac{y\left(y-x\right)}{x}\)