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(x+1)2-(x-1)2-3(x+1)(x-1)
=[(x+1)-(x-1)][(x+1)+(x-1)]-3(x2-1)
=(x+1-x+1)(x+1+x-1)-3x2+3
=2.2x-3x2+3
=-3x2+4x+3
D=(x2+x+1)(x2-x+1)(x4-x2+1)(x8-x4+1)
\(=\left(\left(x^2+1\right)^2-x^2\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(\left(x^4+1\right)^2-x^4\right)\left(x^8-x^4+1\right).\)
\(=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)=\left(x^8+1\right)-x^8=x^{16}+x^8 +1\)
\(x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\)
\(=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\)(1)
Thay \(x=\frac{1}{2};y=-100\) vào (1), ta có:
\(-2.\frac{1}{2}.-100=100\)
\(=\left(x^2-3x+1+3-x-x\right)^2\)
\(=\left(-4x+4\right)^2\)
\(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)\)
\(=2x.2-3x^2+1\)
\(=4x-3x^2+1\)
`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`
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
Ta có : (x+ 2)^2 - 2(x- 1) ( x+ 3) + x^2
= x^2 + 4x + 4 +( - 2x + 2)( x+ 3) + x^2
= x^2 + 4x + 4- 2x^2- 6x + 2x+ 6 + x^2
= 10
\(x^2\left(x-2\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-2x^2-\left(x^3-1\right)\)
\(=x^3-2x^2-x^3+1\)
\(=-2x^2+1\)