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`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.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)
a) \(\left(x+3\right)\left(x-1\right)-x\left(x-5\right)=x^2+2x-3-x^2+5x=7x-3\)
b) \(\left(2x-3\right)\left(2x+3\right)-4\left(x+2\right)^2=4x^2-9-4x^2-16x-16=-16x-25\)
c) \(=x^3-3x^2+3x-1-x^3-8+3x^2=3x-9\)
a) 2x( x - 7 ) - ( x + 3 )( x - 2 ) - ( x + 4 )( x - 4 )
= 2x2 - 14x - ( x2 + x - 6 ) - ( x2 - 16 )
= 2x2 - 14x - x2 - x + 6 - x2 + 16
= 22 - 15x
b) ( 2x + 5 )( x - 2 ) - 3( x + 2 )2 + ( x + 1 )2
= 2x2 + x - 10 - 3( x2 + 4x + 4 ) + x2 + 2x + 1
= 3x2 + 3x - 9 - 3x2 - 12x - 12
= -9x - 21
c) ( x + 3 )( x - 3 ) - ( x + 5 )( x - 1 ) - ( x - 4 )2
= x2 - 9 - ( x2 + 4x - 5 ) - ( x2 - 8x + 16 )
= x2 - 9 - x2 - 4x + 5 - x2 + 8x - 16
= -x2 + 4x - 20
d) 2x( x + 1 )2 - ( x - 1 )3 - ( x - 2 )( x2 + 2x + 4 )
= 2x( x2 + 2x + 1 ) - ( x3 - 3x2 + 3x - 1 ) - ( x3 - 8 )
= 2x3 + 4x2 + 2x - x3 + 3x2 - 3x + 1 - x3 + 8
= 7x2 - x + 9
e) ( x + 5 )( x - 5 )( x + 2 ) - ( x + 2 )3
= ( x2 - 25 )( x + 2 ) - ( x3 + 6x2 + 12x + 8 )
= x3 + 2x2 - 25x - 50 - x3 - 6x2 - 12x - 8
= -4x2 - 37x - 58