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Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
Bài 1.
A = x2 + 2xy + y2 = ( x + y )2 = ( -1 )2 = 1
B = x2 + y2 = ( x2 + 2xy + y2 ) - 2xy = ( x + y )2 - 2xy = (-1)2 - 2.(-12) = 1 + 24 = 25
C = x3 + 3xy( x + y ) + y3 = ( x3 + y3 ) + 3xy( x + y ) = ( x + y )( x2 - xy + y2 ) + 3xy( x + y )
= -1( 25 + 12 ) + 3.(-12).(-1)
= -37 + 36
= -1
D = x3 + y3 = ( x3 + 3x2y + 3xy2 + y3 ) - 3x2y - 3xy2 = ( x + y )3 - 3xy( x + y ) = (-1)3 - 3.(-12).(-1) = -1 - 36 = -37
Bài 2.
M = 3( x2 + y2 ) - 2( x3 + y3 )
= 3( x2 + y2 ) - 2( x + y )( x2 - xy + y2 )
= 3( x2 + y2 ) - 2( x2 - xy + y2 )
= 3x2 + 3y2 - 2x2 + 2xy - 2y2
= x2 + 2xy + y2
= ( x + y )2 = 12 = 1
Bài 1:
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b+\left(a-b\right)\right).\left(a+b-\left(a-b\right)\right)\)
\(=2a.2b\)
\(=4ab\)
Câu 1:
a) (a +b )2 - ( a -b )2
=a2+b2-a2+b2
=2b2
b) (a + b )3- ( a - b )3 - 2b3
=a3+b3-a+b3-2b3
=a3-a
c) ( x+y+z)2 - 2(x+y+z)(x+y) + (x + y )2
=x2+xy+xz+xy+y2+yz+xz+yz+z2-2.(x2+xy+xz+xy+y2+yz)+x2+xy+xy+y2
=x2+y2+z2+2xy+2xz+2yz-2x2-2y2-4xy-2xz-2yz+x2+2xy+y2
=0
a) Ta có: \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)
\(=\left[\left(3-xy^2\right)-\left(2+xy^2\right)\right]\cdot\left[\left(3-xy^2\right)+\left(2+xy^2\right)\right]\)
\(=\left(3-xy^2-2-xy^2\right)\cdot\left(3-xy^2+2+xy^2\right)\)
\(=5\cdot\left(1-2xy^2\right)\)
\(=5-10xy^2\)
b) Ta có: \(9x^2-\left(3x-4\right)^2\)
\(=\left[3x-\left(3x-4\right)\right]\left[3x+\left(3x-4\right)\right]\)
\(=\left(3x-3x+4\right)\cdot\left(3x+3x-4\right)\)
\(=4\cdot\left(6x-4\right)\)
\(=24x-16\)
c) Ta có: \(\left(a-b^2\right)\left(a+b^2\right)\)
\(=a^2-b^4\)
d) Ta có: \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
\(=a^4+4a^3+4a^2-9\)
e) Ta có: \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=x^2+xy-6x-yx-y^2+6y+6x+6y-36\)
\(=x^2-y^2+12y-36\)
f) Ta có: \(\left(y+2z-3\right)\left(y-2z-3\right)\)
\(=\left(y-3\right)^2-\left(2z\right)^2\)
\(=y^2-6y+9-4z^2\)
g) Ta có: \(\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(=\left(2y\right)^3-5^3\)
\(=8y^3-125\)
h) Ta có: \(\left(3y+4\right)\left(9y^2-12y+16\right)\)
\(=\left(3y\right)^3+4^3\)
\(=27y^3+64\)
i) Ta có: \(\left(x-3\right)^3+\left(2-x\right)^3\)
\(=\left(x-3\right)^3-\left(x-2\right)^3\)
\(=x^3-9x^2+27x-27-\left(x^3-6x^2+12x-8\right)\)
\(=x^3-9x^2+27x-27-x^3+6x^2-12x+8\)
\(=-3x^2+15x-19\)
j) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\cdot\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
\(=6x^2y+2y^3\)
a, \(\left(2x+1\right)\left(2x-1\right)\left(x-7\right)=4x^3-28x^2-x+7\)
b, \(\left(3x^2\right)\left(5x+2\right)\left(7x-3\right)=105x^4-3x^3-18x^2\)
a. \(\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\left(x+y\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5\) ( đpcm )
b. \(\left(3-a\right)\left(a^2+3a+9\right)\)
\(=3a^2+9a+27-a^3-3a^2-9a\)
\(=27-a^3\)( đpcm )
Bài 1:
a, \(\left(x-y\right)^2=x^2+y^2+2xy-4xy=\left(x+y\right)^2-4xy\)
Thay \(x+y=3,xy=-4\), ta có:
\(\left(x-y\right)^2=3^2-4.\left(-4\right)=25\)
b, \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
Thay \(x+y=3,xy=-4\),ta có:
\(x^3+y^3=3^3-3.\left(-4\right).3=63\)
c, Giải \(\left\{{}\begin{matrix}x+y=3\\xy=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^3-y^3=65\\x^3-y^3=-65\end{matrix}\right.\)
Bài 1:
\(a,\left(x-y\right)^2=\left(x+y\right)^2-4xy=3^2-4.\left(-4\right)=25\)
\(b,x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3\left[\left(x+y\right)^2-3xy\right]\)
\(=3\left(3^2-3.\left(-4\right)\right)=63\)
\(c,\)\(x+y=3\Rightarrow x=3-y\)
Thay vào xy = -4 ,có :
\(\left(3-y\right)y=-4\Leftrightarrow-y^2+3y+4=0\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-4=-1\\x=3-\left(-1\right)=4\end{matrix}\right.\)
\(TH1:x^3-y^3=\left(4^3\right)-\left(-1\right)^3=65\)
\(TH2:x^3-y^3=\left(-1\right)^3-4^3=-65\)
Bài 2:
\(A=x^2-3x=\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{4}\)\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(Min_A=-\frac{9}{4}\Leftrightarrow x=-\frac{3}{2}\)
\(B=2x^2+x=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)-\frac{1}{8}\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{4}\)
\(Min_B=-\frac{1}{8}\Leftrightarrow x=-\frac{1}{4}\)