Khai triển các biểu thức sau:
a) a 3 + 4 y 2 ; b) 1 x − 3 y 2 ;
c) x 2 − yz 6 x 2 + yz 6 ; d) x 2 + 2 5 y x 2 − 2 5 y .
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a. (x + y)2 = x2 + 2xy + y2
b. (x - 2y)2 = x2 - 4xy - 4x2
c. (xy2 + 1)(xy2 - 1) = x2y4 - 1
d. (x + y)2(x - y)2 = (x2 + 2xy + y2)(x2 - 2xy + y2) = x4 - (2xy + y2)2 = x4 - (4x2y2 + y4) = x4 - 4x2y2 - y4
Chucs hocj toots
Câu 2:
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)
e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)
a) \({\left( {3x + y} \right)^4} = {\left( {3x} \right)^4} + 4.{\left( {3x} \right)^3}y + 6.{\left( {3x} \right)^2}{y^2} + 4.\left( {3x} \right){y^3} + {y^4}\)
\( = 81{x^4} + 108{x^3}y + 54{x^2}{y^2} + 12x{y^3} + {y^4}\)
b) \(\begin{array}{l}{\left( {x - \sqrt 2 } \right)^5} = \left( {x + (-\sqrt 2) } \right)^5 ={x^5} + 5.{x^4}.\left( { - \sqrt 2 } \right) + 10.{x^3}.{\left( { - \sqrt 2 } \right)^2} + 10.{x^2}.{\left( { - \sqrt 2 } \right)^3} + 5.x.{\left( { - \sqrt 2 } \right)^4} + 1.{\left( { - \sqrt 2 } \right)^5}\\ = {x^5} - 5\sqrt 2 .{x^4} + 20{x^3} - 20\sqrt 2 .{x^2} + 20x - 4\sqrt 2 \end{array}\)
a) Áp dụng công thức nhị thức Newton, ta có
\(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\\ = \left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right] + \left[ {{{4.2}^3}.\left( {\sqrt 2 } \right) + 4.2.{{\left( {\sqrt 2 } \right)}^3}} \right]\\ = 68 + 48\sqrt 2 \end{array}\)
b) Áp dụng công thức nhị thức Newton, ta có
\({\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\)
\({\left( {2 - \sqrt 2 } \right)^4} = \left( {2 +(- \sqrt 2 )} \right)^4= {2^4} + {4.2^3}.\left( { - \sqrt 2 } \right) + {6.2^2}.{\left( { - \sqrt 2 } \right)^2} + 4.2.{\left( { - \sqrt 2 } \right)^3} + {\left( { - \sqrt 2 } \right)^4}\)
Từ đó,
\(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4} = 2\left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right]\\ = 2\left( {16 + 48 + 4} \right) = 136\end{array}\)
c) Áp dụng công thức nhị thức Newton, ta có
\(\begin{array}{l}{\left( {1 - \sqrt 3 } \right)^5} = \left( {1 +(- \sqrt 3 )} \right)^5= 1 + 5.\left( { - \sqrt 3 } \right) + 10.{\left( { - \sqrt 3 } \right)^2} + 10.{\left( { - \sqrt 3 } \right)^3} + 5.{\left( { - \sqrt 3 } \right)^4} + 1.{\left( { - \sqrt 3 } \right)^5}\\ = \left[ {1 + 10.{{\left( { - \sqrt 3 } \right)}^2} + 5.{{\left( { - \sqrt 3 } \right)}^4}} \right] + \left[ {5.\left( { - \sqrt 3 } \right) + 10.{{\left( { - \sqrt 3 } \right)}^3} + 1.{{\left( { - \sqrt 3 } \right)}^5}} \right]\\ = 76 - 44\sqrt 3 \end{array}\)
a) \({\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4.{\left( {2x} \right)^3}{.1^1} + 6.{\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4} = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1\)
b) \(\begin{array}{l}{\left( {3y - 4} \right)^4} = {\left[ {3y + \left( { - 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4.{\left( {3y} \right)^3}.\left( { - 4} \right) + 6.{\left( {3y} \right)^2}.{\left( { - 4} \right)^2} + 4.{\left( {3y} \right)^1}{\left( { - 4} \right)^3} + {\left( { - 4} \right)^4}\\ = 81{y^4} - 432{y^3} + 864{y^2} - 768y + 256\end{array}\)
c) \({\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4.{x^3}.{\left( {\frac{1}{2}} \right)^1} + 6.{x^2}.{\left( {\frac{1}{2}} \right)^2} + 4.x.{\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}\)
d) \(\begin{array}{l}{\left( {x - \frac{1}{3}} \right)^4} = {\left[ {x + \left( { - \frac{1}{3}} \right)} \right]^4} = {x^4} + 4.{x^3}.{\left( { - \frac{1}{3}} \right)^1} + 6.{x^2}.{\left( { - \frac{1}{3}} \right)^2} + 4.x.{\left( { - \frac{1}{3}} \right)^3} + {\left( { - \frac{1}{3}} \right)^4}\\ = {x^4} - \frac{4}{3}{x^3} + \frac{2}{3}{x^2} - \frac{4}{27}x + \frac{1}{{81}}\end{array}\)
a) \({\left( {4y - 1} \right)^4} = {\left[ {4y + \left( { - 1} \right)} \right]^4} = 256{y^4} - 256{y^3} + 96{y^2} - 16y + 1\)
b) \({\left( {3x + 4y} \right)^5} = 243{x^5} + 1620{x^4}y + 4320{x^3}{y^2} + 5760{x^2}{y^3} + 3840x{y^4} + 1024{y^5}\)
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !
a) \({\left( {x + 1} \right)^5} = {x^5} + 5.{x^4}.1 + 10.{x^3}{.1^2} + 10.{x^2}{.1^3} + 5.{x^1}{.1^4} +{1^5} = {x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 5x + 1\)
b) \(\begin{array}{l}{\left( {x - 3y} \right)^5} = {\left[ {x + \left( { - 3y} \right)} \right]^5} = {x^5} + 5{x^4}{\left( { - 3y} \right)^1} + 10{x^3}{\left( { - 3y} \right)^2} + 10{x^2}{\left( { - 3y} \right)^3} + 5{x^1}{\left( { - 3y} \right)^4} + {\left( { - 3y} \right)^5}\\ = {x^5} - 15{x^4}y + 90{x^3}{y^2} - 270{x^2}{y^3} + 405x{y^4} - 243{y^5}\end{array}\)
\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)
\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)
a) a 2 9 + 8 3 ay + 16 y 2 . b) 1 x 2 − 6 xy + 9 y 2 .
c) x 2 4 − y 2 z 2 36 . d) x 4 − 4 25 y 2 .