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\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
`#3107`
`a)`
`(6x - 2)^2 + 4(3x - 1)(2 + y) + (y + 2)^2 - (6x + y)^2`
`= [(6x - 2)^2 - (6x + y)^2] + 4(3x - 1)(2 + y) + (2 + y)^2`
`= (6x - 2 - 6x - y)(6x -2 + 6x + y) + (2 + y)*[ 4(3x - 1) + 2 + y]`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x - 4 + 2 + y)`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x + y - 2)`
`= (12x + y - 2)(2 - y + 2 + y)`
`= (12x + y - 2)*4`
`= 48x + 4y - 8`
`b)`
\(5(2x-1)^2+2(x-1)(x+3)-2(5-2x)^2-2x(7x+12)\)
`= 5(4x^2 - 4x + 1) + 2(x^2 + 2x - 3) - 2(25 - 20x + 4x^2) - 14x^2 - 24x`
`= 20x^2 - 20x + 5 + 2x^2 + 4x - 6 - 50 + 40x - 8x^2 - 14x^2 - 24x`
`= - 51`
`c)`
\(2(5x-1)(x^2-5x+1)+(x^2-5x+1)^2+(5x-1)^2-(x^2-1)(x^2+1)\)
`= [ 2(5x - 1) + x^2 - 5x + 1] * (x^2 - 5x + 1) + (5x - 1)^2 - [ (x^2)^2 - 1]`
`= (10x - 2 + x^2 - 5x + 1) * (x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= (x^2 + 5x - 1)(x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= x^4 - (5x - 1)^2 + (5x - 1)^2 - x^4 + 1`
`= 1`
`d)`
\((x^2+4)^2-(x^2+4)(x^2-4)(x^2+16)-8(x-4)(x+4)\)
`= (x^2 + 4)*[x^2 + 4 - (x^2 - 4)(x^2 + 16)] - 8(x^2 - 16)`
`= (x^2 + 4)(x^4 + 12x^2 - 64) - 8x^2 + 128`
`= x^6 + 16x^4 - 16x^2 - 256 - 8x^2 + 128`
`= x^6 + 16x^4 - 24x^2 - 128`
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/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a)-(x-y)(x2+xy-1)=-(x3+x2y-x-x2y-xy2+y)
=-(x3-xy2-x+y)
=-x3+xy2+x-y
b)x2(x-1)-(x3+1)(x-y)=x3-x2-x3+x2y-x+y
=-x2+x2y-x+y
c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x2-3x-4x+2-15x2-10x-3x-2
=-9x2-20x
d) hình như bạn ghi lỗi
Bài 2: C=x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy
=-2xy
Thay x=1/2,y=-1 vào C, ta có:
C=-2.1/2.(-1)=1
Vậy C=1 khi x=1/2 và y=-1.