Viết các biểu thức sau dưới dạng tích:
a) 27 – y 3 ; b) 125 + t 3 ;
c) a 6 + 8 b 3 ; d) z 9 – 27 t 12 .
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a: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
b: \(x^3-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c: \(8x^3+y^3=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
a) \(\left(x+3\right)\cdot\left(x^2-3x+9\right)\)
b) \(\left(x-\dfrac{1}{2}\right)\cdot\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c) \(\left(2x+y\right)\cdot\left(4x^2-2xy+y^2\right)\)
a) \(8y^3+1\)
\(=\left(2y\right)^3+1^3\)
\(=\left(2y+1\right)\left(4y^2-2y+1\right)\)
b) \(y^3-8\)
\(=y^3-2^3\)
\(=\left(y-2\right)\left(y^2+2y+4\right)\)
`8y^3 + 1 = (2y+1)(4y^2 - 2y + 1)`
`y^3 -8 =(y-2)(y^2+2y+4)`
a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)
\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)
\(=x\left(x+2y+3z\right)\)
b: \(\left(x+3\right)^2+4\left(x+3\right)+4\)
\(=\left(x+3+2\right)^2\)
\(=\left(x+5\right)\left(x+5\right)\)
c: \(25+10\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1+5\right)^2\)
\(=\left(x+6\right)\left(x+6\right)\)
\(27-x^3=3^3-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
Bài 1:
c: \(\left(-5x-y\right)^3=-125x^3-75x^2y-15xy^2-y^3\)
h: \(\left(3y-2x^2\right)^3=27y^3-54y^2x^2+36yx^4-8x^6\)
\(a,27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
Các câu còn lại lm tương tự nhé.
hok tốt!
a) \(27-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
b) \(8x^3+0,001=\left(2x+0,1\right)\left(4x^2-0,2x+0,01\right)\)
c) \(\frac{x^3}{125}-\frac{y^3}{27}=\left(\frac{x}{5}-\frac{y}{3}\right)\left(\frac{x^2}{25}+\frac{xy}{15}+\frac{y^2}{9}\right)\)
p/s: chúc bạn học tốt
\(\begin{array}{l}8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}\\ = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.3y + 3.\left( {2x} \right).{\left( {3y} \right)^2} - {\left( {3y} \right)^3}\\ = {\left( {2x - 3y} \right)^3}\end{array}\)
\(a)27{{\rm{x}}^3} + 1 = {\left( {3{\rm{x}}} \right)^3} + 1 = \left( {3{\rm{x}} + 1} \right).\left[ {{{\left( {3{\rm{x}}} \right)}^2} - 3{\rm{x}}.1 + {1^2}} \right] = \left( {3{\rm{x}} + 1} \right)\left( {9{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)\)
\(b)64 - 8{y^3} = {4^3} - {\left( {2y} \right)^3} = \left( {4 - 2y} \right)\left[ {{4^2} + 4.2y + {{\left( {2y} \right)}^2}} \right] = \left( {4 - 2y} \right)\left( {16 + 8y + 4{y^2}} \right)\)
\(8{{\rm{x}}^3} - 36{{\rm{x}}^2}y + 54{\rm{x}}{y^2} - 27{y^3} = {\left( {2{\rm{x}}} \right)^3} - 3.\left( {2{\rm{x}}} \right).3y + 3.2{\rm{x}}.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {2{\rm{x}} - 3y} \right)^3}\)