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a ) \(x^3+1=x^3+1^3=\left(x+1\right)^3-3x\left(x+1\right)\)
b ) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
c ) \(x^3-8=x^3-2^3=\left(x-2\right)^3+3xy\left(x-2\right)\)
d ) \(x-4=\left(\sqrt{x}\right)^2-2^2=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
e ) \(x-1=\left(\sqrt{x}\right)^2-1^2=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
f )\(x\sqrt{x}+1=\left(\sqrt{x}\right)^3+1^3=\left(\sqrt{x}+1\right)-3\sqrt{x}\left(\sqrt{x}+1\right)\)
g ) \(x\sqrt{x}-1=\left(\sqrt{x}\right)^3-1^3=\left(\sqrt{x}-1\right)+3\sqrt{x}\left(\sqrt{x}-1\right)\)
h ) \(8-12x+6x^2-x^3=2^3-3.2^2x+3.2x^2-x^3=\left(2-x\right)^3\)
a) x3 + 1 = ( x + 1 ).( x2 - x + 1 )
b) x3 + y3 = ( x + y ).( x2 - xy + y2 )
c) x3 - 8 = x3 - 23 = ( x - 2 ).( x2 + 2x + 4 )
h) 8 - 12x + 6x2 - x3 = - x3 + 8 + 6x2 - 12x
= -( x3- 8 ) + 6x.( x - 2 )
= -( x - 2 ).( x2 + 2x +4 ) + 6x.( x - 2 )
= -( x - 2 ).( x2 + 2x + 4 + 6x )
= -( x - 2 ).( x2 + 10x + 4 )
= -( x - 2 )3
Bài 1:
a) -16 +(x-3)2
<=> (x-3)2-16
<=> (x-3)2 -42
<=> (x-3-4)(x-3+4)
<=> (x-7)(x+1)
b) 64+16y+y2
<=> y2 + 2.8.y + 82
<=> (y+8)2
c) \(\dfrac{1}{8}-8x^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
d)\(x^2-x+\dfrac{1}{4}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)
e) x4 + 4x2 + 4
<=> (x2)2 + 2.2.x2 +22
<=> (x2 + 2)2
g)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
a) (7x + 4)2 - (7x + 4)(7x - 4)
= 49x2 + 56x + 16 - 49x2 + 16
= 56x + 32
b) (x - 2y)3 - 6xy(x - 2y)
= x3 - 6x2y + 12xy2 - 8y3 - 6x2y + 12xy2
= x3 - 12x2y + 24xy2 - 8y3
c) (3x + y)(9x2 - 3xy + y2) - (3xy)3 - 27x2y
= 27x3 + y3 - (3xy)3 - 27x2y
d) 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2
= 5x2 - 45 + 4x2 + 12x + 9 + x2 - 12x + 36
= 10x2
e) (2x + 3)2 + (2x - 3)2 - 2(4x2 - 9)
= (2x + 3)2 + (2x - 3)2 - 2(2x - 3)(2x + 3)
= (2x + 3 - 2x + 3)2
= 62 = 36
g) (x + 2)3 + (x - 2)3 + x3 - 3x(x - 2)(x + 2)
= (x+2+x-2)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4) + x3 - 3x3 + 12x
= 2x(x2 + 8) + x3 - 3x3 + 12x
= 2x3 + 16x + x3 - 3x3 + 12x
= 28x
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a) Ta có: \(\left(x-3\right)^3\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(=x^3-9x^2+27x^2-27\)
b) Ta có: \(\left(2x-3\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)
\(=8x^3-36x^2+54x-27\)
c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)
\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
d) Ta có: \(\left(x^2-2\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)
\(=x^6-6x^4+12x^2-8\)
e) Ta có: \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-24x^2y+36xy^2-27y^3\)
f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)
\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)
a) Ta có: \(\left(x+1\right)^3\)
\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b) Ta có: \(\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+3\cdot4x^2\cdot3+27\cdot2x+27\)
\(=8x^3+36x^2+54x+27\)
c) Ta có: \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+2\cdot x^2\cdot\frac{1}{2}+2\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=x^3+x^2+\frac{1}{2}x+\frac{1}{8}\)
d) Ta có: \(\left(x^2+2\right)^3\)
\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3\)
\(=x^6+6x^4+12x^2+8\)
e) Ta có: \(\left(2x+3y\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2+\left(3y\right)^3\)
\(=8x^3+36x^2y+54xy^2+27y^3\)
f) Ta có: \(\left(\frac{1}{2}x+y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3+3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)
\(=\frac{1}{8}x^3+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)
a) \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
b) \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
c) \(x^3-y^3=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
d) \(x^2-4=\left(x-2\right)\left(x+2\right)\)
e) \(x^3+y^3=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
x3 - 1 = ( x - 1 ).( x2 + x + 1 )
x2 - y2 = ( x - y ).( x + y )
x3 - y3 = ( x - y ).( x2 + xy + y2 )
x2 - 4 = x2 - (2)2 = ( x - 2 ).( x + 2 )
x3 + y3 = ( x + y ).( x2 - xy + y2 )