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\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)
A = ( 2x + 1 ).( x - 2) - x.( 2x - 3 ) + 7
A = 2x.x + 2x.(-2) + 1x + 1.(-2) - x.2x + x.(-3) + 7
A = \(2x^2\) - 4x + x - 2 - \(2x^2\) - 3x + 7
A = 6x + 5
ko biết đúg k ạk
B = \(\left(x-1\right)^2\) - \(2\left(x^2-1\right)\)+ \(\left(x+1\right)^2\)
B = \(x^2\)- 2x.1 + \(1^2\) - \(2.x^2\) - 2.1 + \(x^2\)+ 2x.1 + \(1^2\)
B = \(x^2\)- 2x + 1 - \(2x^2\) - 2 + \(x^2\)+ 2x + 1
B = \(2x^2\) - 2x + 2 - \(2x^2\) + 1
B = - 2x
>< k bt đúg k -.-
`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`
a) (2x+1)^2+2(4x^2-2)+(2x-1)^2=4x2+4x+1+8x2-4+4x2-4x+1=16x2-2
( x - 3 ) ( x2 + 3x + 9 ) - x ( x2 - 2 ) - 2 ( x - 1 )
= x3 - 27 - x3 + 2x - 2x + 2
= - 25
cách1) nhân vào rùi rút gọn, tự làm nhé, học toán là z
cách2) A = (x-1)(x5 + x4 + x3 ) + (x-1)( x2 +x + 1)
A = x3(x3 -1) + (x3 -1) = (x3-1)(x3 +1)
A = x6 - 1
(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
\(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)\)
\(=2x.2-3x^2+1\)
\(=4x-3x^2+1\)