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a) \(27x^3+8^3\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left[\left(3x\right)^2+6x+2^2\right]\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
b) \(8x^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
c) \(x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(27x^3+8\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(x^2+4xy+4y^2\)
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
_Minh ngụy_
a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)
Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))
\(\frac{1}{4}a^2+2ab+4b^2\)
\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a+2b\right)^2\)
b. \(25+10x+x^2\)
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)
\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)
\(=\left(y^4-\frac{1}{3}\right)^2\)
a) \(27{x^3} + {y^3} = {\left( {3x} \right)^3} + {y^3} = \left( {3x + y} \right)\left( {9{x^2} - 3xy + {y^2}} \right)\);
b) \({x^3} - 8{y^3} = {x^3} - {\left( {2y} \right)^3} = \left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right)\).
a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
1) \(2x^2-5x+3=2x^2-2x-3x+3=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)=\left(2x+2-5\right)\left(x+1-2\right)=\left(2\left(x+1\right)-5\right)\left(x+1-2\right)\)
\(=\left(2y-5\right)\left(y-2\right)\)
\(a)27{{\rm{x}}^3} + 1 = {\left( {3{\rm{x}}} \right)^3} + 1 = \left( {3{\rm{x}} + 1} \right).\left[ {{{\left( {3{\rm{x}}} \right)}^2} - 3{\rm{x}}.1 + {1^2}} \right] = \left( {3{\rm{x}} + 1} \right)\left( {9{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)\)
\(b)64 - 8{y^3} = {4^3} - {\left( {2y} \right)^3} = \left( {4 - 2y} \right)\left[ {{4^2} + 4.2y + {{\left( {2y} \right)}^2}} \right] = \left( {4 - 2y} \right)\left( {16 + 8y + 4{y^2}} \right)\)
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)`