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1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)\)
\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-8x^2+20x-8\)
\(=-2x^2+3x-3\)
a) x(x+4)(x-4)-(x2+1)(x2-1)
=>x(x2-42)-(x4-12)
=>x3-16x-x4+1
=>-x4-x3-15x
b) 7x(4y-x)+4y(y-7x)-2(2y2-3.5x)
=>28xy-7x2+4y2-28xy-4y2+30x
=>-7x2+30x
c) (3x+1)(2x-5)-4(2x2-5x+2)
=>6x2-15x+2x-5-8x2+20x-8
=>-2x2+7x-13
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x=t\)
\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)
\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)
Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
b: Ta có: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1-3ab+3ab\)
=1
a) -4x2 + 8x - 4
= - (4x2 - 8x + 4)
= - (2x - 2)2
b) -x52 + 10 x - 5
= - 5(x2 - 2x + 1)
= - 5(x - 1)2
`4(x-6)-x^2 (2+3x)+x(5x-4)+3x^2 (x-1)`
`=4x-24-2x^2 -3x^3 +5x^2-4x+3x^3-3x^2`
`=-24`
\(4\left(x-6\right)-2x\left(2+3x\right)+x\left(5x-4\right)+3x2\left(x-1\right)\\ =4x-24-4x-6x^2+5x^2-4x+6x^2+6x\\ =2x+5x^2-24\)
\(M=2x\left(-3x+2x^3\right)-x^2\left(3x^2-2\right)-x^2\left(x^2-4\right)\)
\(=-6x^2+4x^4-3x^4+2x^2-x^4+4x^2\)
\(=0\)
Đặt \(A=75\left(4^{2017}+4^{2016}+4^{2015}+...+4^2+5\right)+25\)
\(B=4^{2017}+4^{2016}+4^{2015}+...+4^2+5\)
\(=4^{2017}+4^{2016}+4^{2015}+...+4^2+4+1\)
\(\Rightarrow4B=4^{2018}+4^{2017}+4^{2016}+...+4^3+4^2+4\)
\(\Rightarrow4B-B=\left(4^{2018}+4^{2017}+4^{2016}+...+4^3+4^2+4\right)-\left(4^{2017}+4^{2016}+4^{2015}+...+4^2+4+1\right)\)
\(3B=4^{2018}+4^{2017}+4^{2016}+...+4^3+4^3+4-4^{2017}-4^{2016}-4^{2015}-...-4^2-4-1\)
\(3B=4^{2018}-1\)
\(\Rightarrow B=\dfrac{4^{2018}-1}{3}\)
Suy ra: \(A=75.\dfrac{4^{2018}-1}{3}+25\)
\(A=25.\left(4^{2018}-1\right)+25\)
\(=25\left(4^{2018}-1+1\right)\)
\(=25.4^{2018}\)