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(2x-1).(2x+1).(4x^2+1)
= (4x^2 - 1).(4x^2+1)
= (4x^2)^2 - 1
= 16x^4 - 1
a) \(12\left(2x-5\right)^2-3\left(1+4x\right)\left(4x-1\right)\)
\(=12\left[\left(2x\right)^2-2.2x.5+5^2\right]-3\left(4x+1\right)\left(4x-1\right)\)
\(=12\left(4x^2-20x+25\right)-3\left[\left(4x\right)^2-1\right]\)
\(=48x^2-240x+300-3\left(16x^2-1\right)\)
\(=48x^2-240x+300-48x^2+3\)
\(=-240x+303\)
\(4x^2-25y^2\)
\(\left(2x\right)^2-\left(5y\right)^2\)
\(\left(2x-5y\right)\left(2x+5y\right)\)
chọn c
\(27x^3-1\) =\(\left(3x\right)^3-1^3\)
Nhớ tick và theo dõi mình nha!
a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)
\(=\left(2x\right)^3-1=8x^3-1\)
b,\(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)
`a)(2x-1)(4x^2+2x+1)`
`=(2x-1)[(2x)^2+2x.1+1^2]`
`=(2x)^3-1^3`
`=8x^3-1`
Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`
`b)(x+2y+z)(x+2y-z)`
`=[(x+2y)+z][(x+2y)-z]`
`=(x+2y)^2-z^2`
`=x^2+2.x.2y+(2y)^2-z^2`
`=x^2+4xy+4y^2-z^2`
Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`
`(A+B)^2=A^2+2AB+B^2`
\(\left(5x+3yz\right)^2=25x^2+30xyz+9y^2z^2\\ \left(2x-3\right)^3=8x^3-36x^2+54x-27\)
\(\left(y^{2x}+3yz\right)^2=y^{4x}+6y^{2x+1}z+9y^2z^2\\ \left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\\ \left(y-5\right)\left(25+2y+y^2+3y\right)=\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
(4x - 5)^2 = (4x)^2 - 2.4x.(-5) + (5)^2 = 16x^2 - 13x + 25
(4x-5)^2
= ( 4x)^2 - 2 * 4x * 5 + 5^2
= 16x^2 - 40x + 25