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`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
Answer:
\(\left(2x+1\right)^2+\left(2x-1\right)^2-2\left(1+2x\right)\left(2x-1\right)\)
\(=(4x^2+4x+1)+(4x^2-4x+1)-2(4x^2-1)\)
\(=4x^2+4x+1+4x^2-4x+1-8x^2+2\)
\(=(4x^2+4x^2-8x^2)+(4x-4x)+(1+1+2)\)
\(=4\)
\((x-1)^3-(x+2)(x^2-2x+4)+3(x-1)(x+1)\)
\(=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-1)\)
\(=x^3-3x^2+3x-1-x^3-8+3x^2-3\)
\(=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-3)\)
\(=3x-12\)
(a) Điều kiện : \(x\ne-1.\)
Ta có : \(P=\dfrac{x^4+x}{x^2-x+1}+1-\dfrac{2x^2+3x+1}{x+1}\)
\(=\dfrac{x\left(x^3+1\right)}{x^2-x+1}+1-\dfrac{\left(2x+1\right)\left(x+1\right)}{x+1}\)
\(=\dfrac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}+1-\left(2x+1\right)\)
\(=x\left(x+1\right)+1-2x-1\)
\(=x^2-x.\)
Vậy : Với mọi \(x\ne-1\) thì \(P=x^2-x.\)
(b) Ta có : \(P=x^2-x\)
\(=\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy : \(MinP=-\dfrac{1}{4}.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=\dfrac{1}{2}.\)
(3x - 1)2 + (x + 3)(2x - 1)
= 9x2 - 6x + 1 + 2x2 - x + 6x - 3
= 11x2 - x - 2
(x - 2)(x2 + 2x + 4) - x(x2 - 2)
= x3 - 8 - x3 + 2x
= 2x - 8
b) B = ( x - 2)(x2 + 2x + 4) - x ( x2 -2 )
= x3 - 8 - x3 + 2x
= 2x - 8
( x - 2 )( x2 - 2x + 4 )( x + 2 )( x2 + 2x + 4 )
= [ ( x - 2 )( x2 + 2x + 4 ) ][ ( x + 2 )( x2 - 2x + 4 ) ]
= ( x3 - 8 )( x3 + 8 )
= ( x3 )2 - 82
= x6 - 64
b/ \(4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)
= \(4\left(x^2-1\right)-5x^2+10x+x^2\)
= \(4x^2-4-5x^2+10x+x^2\)
= \(10x-4\)
= \(2\left(5x-2\right)\)
c/ \(\left(3-2x\right)\left(x-2\right)+4\left(x-1\right)\left(x-3\right)-2\left(x-2\right)\left(x+2\right)\)
= \(3x-6-2x^2+4x+4\left(x^2-4x-4\right)-2\left(x^2-4\right)\)
= \(3x-6-2x^2+4x+4x^2-16x-16-8x^2-18\)
= \(-9x-12\)
= \(-3\left(3x+4\right)\)
a/\(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+1-8x^2+24x-18+4\)
\(=-4x^2+20x-13\)
b/ \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2x^2-2y^2+2x^2+2y^2\)
\(=4x^2\)
chúc bạn học tốt
Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-1\right)-x+2\)
\(=x^3-8-x^3+x-x+2\)
\(=-6\)