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\(a,\left(2x+y\right)^2+\left(x-y\right)^2-5\left(x+y\right)\left(x-y\right)\)
\(=4x^2+4xy+y^2+x^2-2xy+y^2-5\left(x^2-y^2\right)\)
\(=4x^2+4xy+y^2+x^2-2xy+y^2-5x^2+5y^2\)
\(=7y^2+2xy\)
a)\(\left(x-2\right)\left(x+2\right)+4\left(1-x^2\right)\)
\(=x^2-4+4-4x^2\)
\(=-3x^2\)
b) \(\left(2x-y\right)^2+\left(2x+y\right)^2-2\left(4x^2-y^2\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2\)
\(=\left(2x-y-2x-y\right)^2\)
\(=\left(-2y\right)^2\)
\(=4y^2\)
bài 1.
a.\(A=x^2-2xy+y^2+x^2+2xy+y^2=2\left(x^2+y^2\right)\)
b.\(B=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)=4xy\)
c.\(C=4a^2+4ab+b^2-\left(4a^2-4ab+b^2\right)=8ab\)
d.\(D=4x^2-4x+1-2\left(4x^2-12x+9\right)+4=-4x^2+20x-13\)
.bài 2
a.\(A=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)=10x+16;x=-\frac{1}{2}\Rightarrow A=9\)
b.\(B=9x^2+24x+16-x^2+16-10x=8x^2+14x+32\Rightarrow x=-\frac{1}{10}\Rightarrow B=\frac{767}{25}\)
c.\(C=x^2+2x+1-\left(4x^2-4x+1\right)+3\left(x^2-4\right)=6x-12\Rightarrow x=1\Rightarrow C=-6\)
d.\(D=x^2-9+x^2-4x+4-2x^2+8x=4x-5\Rightarrow x=-1\Rightarrow A=-9\)
Trả lời:
Bài 1: Rút gọn biểu thức:
a) A = ( x - y )2 + ( x + y )2
= x2 - 2xy + y2 + x2 + 2xy + y2
= 2x2 + 2y2
b) B = ( x + y )2 - ( x - y )2
= x2 + 2xy + y2 - ( x2 - 2xy + y2 )
= x2 + 2xy + y2 - x2 + 2xy - y2
= 4xy
c) C = ( 2a + b )2 - ( 2a - b )2
= 4a2 + 4ab + b2 - ( 4a2 - 4ab + b2 )
= 4a2 + 4ab + b2 - 4a2 + 4ab - b2
= 8ab
d) D = ( 2x - 1 )2 - 2 ( 2x - 3 )2 + 4
= 4x2 - 4x + 1 - 2 ( 4x2 - 12x + 9 ) + 4
= 4x2 - 4x + 1 - 8x2 + 24x - 18 + 4
= - 4x2 + 20x - 13
Bài 2: Rút gọn rồi tính giá trị biểu thức:
a) A = ( x + 3 )2 + ( x - 3 )( x + 3 ) - 2 ( x + 2 )( x - 4 )
= x2 + 6x + 9 + x2 - 9 - 2 ( x2 - 2x - 8 )
= 2x2 + 6x - 2x2 + 4x + 16
= 10x + 16
Thay x = 1/2 vào A, ta có:
\(A=10.\left(-\frac{1}{2}\right)+16=-5+16=11\)
b) B = ( 3x + 4 )2 - ( x - 4 )( x + 4 ) - 10x
= 9x2 + 24x + 16 - x2 + 16 - 10x
= 8x2 + 14x + 32
Thay x = - 1/10 vào B, ta có:
\(B=8.\left(-\frac{1}{10}\right)^2+14.\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)
c) C = ( x + 1 )2 - ( 2x - 1 )2 + 3 ( x - 2 )( x + 2 )
= x2 + 2x + 1 - 4x2 + 4x - 1 + 3 ( x2 - 4 )
= - 3x2 + 6x + 3x2 - 12
= 6x - 12
Thay x = 1 vào C, ta có:
\(C=6.1-12=-6\)
d) D = ( x - 3 )( x + 3 ) + ( x - 2 )2 - 2x ( x - 4 )
= x2 - 9 + x2 - 4x + 4 - 2x2 + 8x
= 4x - 5
Thay x = - 1 vào D, ta có:
\(D=4.\left(-1\right)-5=-9\)
a) \(A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(=x^2-2xy+y^2+x^2+2xy+y^2\)
\(=\left(x^2+x^2\right)-\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(=2x^2+2y^2\)
\(=2.\left(x^2+y^2\right)\)
b) \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)
\(=8ab\)\
c) \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=\left(x^2-x^2\right)+\left(2xy+2xy\right)+\left(y^2-y^2\right)\)
\(=4xy\)
d) \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+1-8x^2+24x-18+4\)
\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)
\(=-4x^2+20x-13\)
\(=-4x^2+20x-25+12\)
\(=-\left(4x^2-20x+25\right)-8\)
\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)
\(=-\left(2x-5\right)^2-8\)
Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
\(2.A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\\ Thayx=\frac{1}{2};y=-100vàoAđược:A=-2.\frac{1}{2}.\left(-100\right)=100\)
\(3.x\left(5-2x\right)+2x\left(x-1\right)=15\Leftrightarrow5x-2x^2+2x^2-2x=15\Leftrightarrow3x=15\Leftrightarrow x=5\)
a) \(\left(x^2+2x+1\right)=x^2+2.x.1+1^2=\left(x+1\right)^2:\left(x+1\right)=x+1\)
b) \(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y2x=4xy\)
a) \(\left(x^2+2x+1\right):\left(x+1\right)=\left(x+1\right)\)
b)\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=4xy\)