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Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
a) \(\left(2a-3\right)\left(a+1\right)-\left(a^2+6a+9\right):\left(a+3\right)\)
\(=\left(2a^2+2a-3a-3\right)-\left(a+3\right)^2:\left(a+3\right)\)
\(=2a^2-a-3-\left(a+3\right)\)
\(=2a^2-a-3-a-3\)
\(=2a^2-2a-6\)
b) \(\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\)
\(=\left(3x-5y\right)\cdot x^2y^2-3x^2y^2+4x^2y^3\)
\(=3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\)
\(=3x^3y^2-x^2y^3-3x^2y^2\)
c) \(x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\)
\(=x\left(x^2-4x+4\right)-\left(x^3+8\right)+4x^2\)
\(=x^3-4x^2+4x-x^3-8+4x^2\)
\(=\left(x^3-x^3\right)+\left(-4x^2+4x^2\right)+4x-8\)
\(=4x-8\)
`a)x^2(x+4)(x-4)-(x^2+1)(x^2-1)`
`=x^2(x^2-16)-(x^2+1)(x^2-1)`
`=x^4-16x^2-(x^4-1)`
`=-16x^2+1`
`b) (a-b+c)^2-(a-c)^2-2ac+2ab`
`=a^2+b^2+c^2-2ab-2bc+2ac-(a^2-2ac+c^2)-2ac+2ab`
`=a^2+b^2+c^2-2ab-2bc+2ac-a^2+2ac-c^2-2ac+2ab`
`=b^2-2bc+2ac`
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
\(a.\left(3x-1\right)^2+\left(x+3\right)\left(2x-1\right)\)
\(=9x^2-6x+1-2x^2+x-6x+3\)
\(=7x^2-11x+4\)
a) (x+2)2+x(x-4)
=x2+4x+4+x2-4x
=2x2+4
b)(x-3)2-(x+3)(x-4)
=x2-6x+9-x2+4x-3x+12
=-5x+12
c) (3x+1)2+3x(2-4x)
=9x2+6x+1+6x-12x2
=-3x2+12x+1
d) (2x-4y)2-(2x-3)(2x-3y)
=4x2-16xy+16y2-4x2+6xy+6x-9y
=16y2-10xy+6x-9y
a: Ta có: \(A=\left(x+2\right)\left(x-4\right)+\left(x+1\right)\left(x-6\right)\)
\(=x^2-4x+2x-8+x^2-6x+x-6\)
\(=2x^2-7x-14\)
b: \(B=\left(2a-b\right)\left(4a^2+2ab+b^2\right)=8a^3-b^3\)
c: \(C=\left(2+x\right)\left(2-x\right)\left(x+4\right)\)
\(=\left(4-x^2\right)\left(x+4\right)\)
\(=4x+16-x^3-4x^2\)