Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
a) \(25{{\rm{x}}^2} - 16 = {\left( {5{\rm{x}}} \right)^2} - {4^2} = \left( {5{\rm{x}} + 4} \right)\left( {5{\rm{x}} - 4} \right)\)
b) \(8{{\rm{x}}^3} + 1 = {\left( {2{\rm{x}}} \right)^3} + {1^3} = \left( {2{\rm{x}} + 1} \right)\left( {4{{\rm{x}}^2} - 2{\rm{x}} + 1} \right)\)
c) \(8{{\rm{x}}^3} - 125 = {\left( {2{\rm{x}}} \right)^3} - {5^3} = \left( {2{\rm{x}} - 5} \right)\left( {4{{\rm{x}}^2} + 10{\rm{x + }}25} \right)\)
d) \(27{{\rm{x}}^3} - {y^3} = {\left( {3x} \right)^3} - {y^3} = \left( {3{\rm{x}} - y} \right)\left( {9{{\rm{x}}^2} + 3{\rm{x}}y + {y^2}} \right)\)
e) \(16{{\rm{a}}^2} - 9{b^2} = {\left( {4{\rm{a}}} \right)^2} - {\left( {3b} \right)^2} = \left( {4{\rm{a}} - 3b} \right)\left( {4{\rm{a}} + 3b} \right)\)
g) \(125{{\rm{x}}^3} + 27{y^3} = {\left( {5{\rm{x}}} \right)^3} + {\left( {3y} \right)^3} = \left( {5{\rm{x}} + 3y} \right)\left( {25{{\rm{x}}^2} - 15{\rm{x}}y + 9{y^2}} \right)\)
Đã trả lời: Câu hỏi của Naryu Wikashi - Toán lớp 8 | Học trực tuyến
a. x2 - 6x + 9
= x2 - 2x3 + 32
= (x - 3)2
b. x2 + x + \(\frac{1}{4}\)
= x2 + 2x\(\frac{1}{2}\)+ \(\left(\frac{1}{2}\right)^2\)
= (x + \(\frac{1}{2}\))2
c. 4x2 - \(\frac{1}{16}\)
= (2x)2 - \(\left(\frac{1}{4}\right)^2\)
= (2x +\(\frac{1}{4}\))(2x - \(\frac{1}{4}\))
d. (a + b)2 - 4
= (a + b)2 - 22
= (a + b + 2)(a + b - 2)
e. (a2 + 9)2 - 36a2
= (a2 + 9)2 - (6a)2
= (a2 + 9 + 6a)(a2 + 9 - 6a)
Tích mình đi
Ai tích sẽ có lợi
vì khi có lợi bạn sẽ được người khác tích lại.
THANKS
\(a){x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}} \\= {x^2} + 2.x.\dfrac{1}{4} + {\left( {\dfrac{1}{4}} \right)^2} \\= {\left( {x + \dfrac{1}{4}} \right)^2}\)
\(b)25{{\rm{x}}^2} - 10{\rm{x}}y + {y^2} \\= {\left( {5{\rm{x}}} \right)^2} - 2.5{\rm{x}}.y + {y^2} \\= {\left( {5{\rm{x}} - y} \right)^2}\)
\(\begin{array}{l}c){x^3} + 9{{\rm{x}}^2}y + 27{\rm{x}}{y^2} + 27{y^3}\\ = {x^3} + 3{{\rm{x}}^2}.3y + 3.x.{\left( {3y} \right)^2} + {\left( {3y} \right)^3}\\ = {\left( {x + 3y} \right)^3}\end{array}\)
\(\begin{array}{l}d)64{{\rm{x}}^3} - 48{{\rm{x}}^2}y + 12{\rm{x}}{y^2} - {y^3}\\ = {\left( {4{\rm{x}}} \right)^3} - 3.{\left( {4{\rm{x}}} \right)^2}.y + 3.4{\rm{x}}.{y^2} - {y^3}\\ = {\left( {4{\rm{x}} - y} \right)^3}\end{array}\)
\(a)27{{\rm{x}}^3} + 1 = {\left( {3{\rm{x}}} \right)^3} + 1 = \left( {3{\rm{x}} + 1} \right).\left[ {{{\left( {3{\rm{x}}} \right)}^2} - 3{\rm{x}}.1 + {1^2}} \right] = \left( {3{\rm{x}} + 1} \right)\left( {9{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)\)
\(b)64 - 8{y^3} = {4^3} - {\left( {2y} \right)^3} = \left( {4 - 2y} \right)\left[ {{4^2} + 4.2y + {{\left( {2y} \right)}^2}} \right] = \left( {4 - 2y} \right)\left( {16 + 8y + 4{y^2}} \right)\)
\(8{{\rm{x}}^3} - 36{{\rm{x}}^2}y + 54{\rm{x}}{y^2} - 27{y^3} = {\left( {2{\rm{x}}} \right)^3} - 3.\left( {2{\rm{x}}} \right).3y + 3.2{\rm{x}}.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {2{\rm{x}} - 3y} \right)^3}\)
a) \(x^2-6x+9=\left(x-3\right)^2\)
b) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
c) \(4x^2-\frac{1}{16}=\left(2x-\frac{1}{4}\right)\left(2x+\frac{1}{4}\right)\)
d) \(\left(a+b\right)^2-4=\left(a+b-2\right)\left(a+b+2\right)\)
e) \(\left(a^2+9\right)^2-36a^2=\left(a^2-6a+9\right)\left(a^2+6a+9\right)\)
\(=\left(a-3\right)^2\cdot\left(a+3\right)^2\)
a) Ta có:
(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2
=x^2 - y^2 - 2yz - z^2.
b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2
=x^2 +2xz- y^2 +z^2.
c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)
= -16 + x^2 - 6x + 9
= x^2 - 6x - 7.
\(a,\left(x+y+z\right)\left(x-y-z\right)\)
\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)
\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)
\(=x^2-y^2-2yz-z^2\)
\(b,\left(x-y+z\right)\left(x+y+z\right)\)
\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)
\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)
\(=x^2+2xz-y^2+z^2\)
\(c,-16+\left(x-3\right)^2\)
\(=-16+x^2-6x+9\)
\(=x^2-6x-7\)
a) \(9{{\rm{x}}^2} - 16 = {\left( {3{\rm{x}}} \right)^2} - {4^2} = \left( {3{\rm{x}} - 4} \right)\left( {3{\rm{x}} + 4} \right)\)
b) \(25 - 16{y^2} = {5^2} - {\left( {4y} \right)^2} = \left( {5 - 4y} \right)\left( {5 + 4y} \right)\)