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a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
a) -4x2 .( 5x - 3 ) = -20x3 + 12x2
b) ( 2x^2 - x +5 ) (4 - 3x ) = 8x^2 - 4x +20 - 6x^3 +3x^2 -15x
= 2x^2 + 3x^2 - 19x +20
c ) ( -3y +x ) ( 2x -y +5) = -6xy + 3y^2 - 15 + 2x^2 - xy +5x
= -7xy + 3y^2 + 2x^2 -15 + 5x
d) nhân -2/3xy với từng hạng tử = - 4x3y - 14/3 x2y2 + 5/6 xy
Bài 2
a) x^2 - 1 = x2 - 12 = ( x+y ) ( x-y )
b) x2 - 16 = x2 - 42 = (x +4) .( x - 4 )
c) 4x2 - 9 = ( 2x )2 - 32 = (2x +3 ) .( 2x-3 )
d) 25/16 - x2 = ( 5/4 )2 - x2 = ( 5/4 +x ) . ( 5/4 - x )
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
bn nên vt thành phân thức thì mọi người sẽ dễ nhìn và sẽ giải giúp bn!!!
Bài 1:
\(\left(3x+4\right)^2=9x^2+24x+16\)
\(\left(x-1\right)^2=x^2-2x+1\)
\(\left(x-3\right)^2=x^2-6x+9\)
\(\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25\)
\(x^2-1=\left(x+1\right)\left(x-1\right)\)
\(x^2-y^2=\left(x+y\right)\left(x-y\right)\)
\(x^2-2=\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)
\(4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)\)
Bài 3:
\(x^2-2x+1=\left(x-1\right)^2\)
\(x^2-10x+25=\left(x-5\right)^2\)
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
\(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25\)