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a) ( x - 3 ) ( x + 7 ) - ( x + 5 ) ( x - 1 )
= ( x2 +7x - 3x - 21 ) - ( x2 -x + 5x - 5 )
= x2 + 4x -21 - x2 - 4x +5
= -16
a: Ta có: \(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)\)
\(=x^2+4x-21-x^2-4x+5\)
=-16
`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`
\(a,=5x-10+2x+6=7x-4\\ b,=x^2+2x+1-x^2+3x+10=5x+11\\ c,=x^2-49-x^2+1=-48\\ d,\text{Đề có sai ko vậy?}\)
`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`
`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`
`=5x^2-5x^2-4x+6x+1+9+245`
`=2x+255`
`b)(x-2)(x^2+2x+4)-(25+x^3)`
`=x^3-8-x^3-25=-33`
Lời giải:
a.
$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$
$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$
$=5x^2+2x+10-(5x^2-245)=2x+255$
b.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x-7\)
\(=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x-7\)
\(=2x^2-7x-15-2x^2+6x+x-7\)
\(=-22\)
a) \(3x\left(x-3\right)-5x\left(x+7\right)\)
\(=3x^2-9x-5x^2-35x\)
\(=-2x^2-44x\)
b) \(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)-12\)
\(=2x^2-3x-2x^2+10x-12\)
\(=7x-12\)
=x2+4x-21-(x2+4x-5)
=x2+4x-21-x2-4x+5
=-16
\(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)=x^2+7x-3x-21-x^2+x-5x+5\)
\(=-16\)