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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`
\(=\left(x^2-3x+1+3-x-x\right)^2\)
\(=\left(-4x+4\right)^2\)
(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
x.(22 - 3 ) - x2 (5x + 1) + x2 = 4x - 3x - 5x3 - x2 + x2 = -5x3 + x
\(A=x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-\left(5x^3+x^2\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
\(=-3x\left(x^2+1\right)\).
x(2x2 - 3) - x2(5x + 1) + x2 = 2x3 - 3x - 5x3 - x2 + x2 = -3x3 - 3x = 3x(-x2 - 1)
\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)
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