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`# \text {04th5}`
`a.`
`P = (5x^2 - 2xy + y^2) - (x^2 + y^2) - (4x^2 - 5xy + 1)`
`= 5x^2 - 2xy + y^2 - x^2 - y^2 - 4x^2 + 5xy - 1`
`= (5x^2 - x^2 - 4x^2) + (-2xy + 5xy) + (y^2 - y^2) - 1`
`= 3xy - 1`
`b.`
\((x^2-5x+4)(2x+3)-(2x^2-x-10)(x-3)\)
`= x^2(2x + 3) - 5x(2x + 3) + 4(2x + 3) - [ 2x^2(x - 3) - x(x - 3) - 10(x - 3)]`
`= 2x^3 + 3x^2 - 10x^2 - 15x + 8x + 12 - (2x^3 - 6x^2 - x^2 + 3x - 19x + 30)`
`= 2x^3 -7x^2 - 7x + 12 - (2x^3 - 7x^2 - 7x + 30)`
`= 2x^3 - 7x^2 - 7x + 12 - 2x^3 + 7x^2 + 7x -30`
`= -30`
Vậy, giá trị của biểu thức không phụ thuộc vào giá trị của biến.
a, \(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)
b, = \(y^2-4x^2+4x^2=y^2\)
Thay y = 10 vào BT trên, ta có:
\(y^2=10^2=100\)
Vậy giá trị của BT là 100
a: \(N=\left(2x-3y\right)\left(2x+3y\right)=\left(2x\right)^2-\left(3y\right)^2\)
\(=4x^2-9y^2\)
Thay x=1/2 và y=1/3 vào N, ta được:
\(N=4\cdot\left(\dfrac{1}{2}\right)^2-9\left(\dfrac{1}{3}\right)^2\)
\(=4\cdot\dfrac{1}{4}-9\cdot\dfrac{1}{9}\)
=1-1
=0
b: \(N=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=\left(2x\right)^3-y^3=8x^3-y^3\)
Khi x=1 và y=3 thì \(N=8\cdot1^3-3^3=8-27=-19\)
Bài 1:
\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3-9x^2y+xy^2-3y^3+5x^2y-15xy^2=3x^3-3y^3-4x^2y-14xy^2\)
Bài 2:
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=x^2+16x+64-2x^2-12x+32+x^2-4x+4=100\\ c,=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)
b: \(=7\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(7-x+y\right)=0\)
a) \(A=\left(x+2\right)\left(x^2-2x+4\right)-x^3+2\)
\(A=x^3+8-x^3+2\)
\(A=10\)
b) \(B=\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(B=x^3-1-\left(x^3+1\right)\)
\(B=x^3-1-x^3-1\)
\(B=-2\)
c) \(C=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(y-3x\right)\left(y^2+3xy+9x^2\right)\)
\(C=\left(2x\right)^3-y^3+y^3-\left(3x\right)^3\)
\(C=8x^3-y^3+y^3-27x^3\)
\(C=-19x^3\)
a)
\(A=\left(x+2\right)\left(x-2\right)\left(x-2\right)-x^3+2\\ =\left(x^2-4\right)\left(x-2\right)-x^3+2\\ =x^3-2x^2-4x+8-x^3+2\\ =-2x^2-4x+10\)
b)
\(B=x^3-1-\left(x^3+1\right)\\ =x^3-1-x^3-1\\ =-2\)
c)
\(C=\left(2x\right)^3-y^3+\left(y\right)^3-\left(3x\right)^3\\ =8x^3-y^3+y^3-27x^3\\ =-19x^3\)
a) (x + 3)(x2 – 3x + 9) – (54 + x3)
= ( x + 3)(x2 – 3.x + 32) – (54 + x3)
= x3 + 33 – (54 + x3)
= x3 + 27 – 54 – x3
= -27
b) (2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2x.y + y2] – (2x – y)[(2x)2 + 2x.y + y2]
= [(2x)3 + y3] – [(2x)3 – y3]
= (2x)3 + y3 – (2x)3 + y3
= 2y3
a) (x + 3)(x2 – 3x + 9) – (54 + x3)
= ( x + 3)(x2 – 3.x + 32) – (54 + x3)
= x3 + 33 – (54 + x3) = x3 + 27 – 54 – x3
= -27
b) (2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2x.y + y2] – (2x – y)[(2x)2 + 2x.y + y2]
= [(2x)3 + y3] – [(2x)3 – y3]
= (2x)3 + y3 – (2x)3 + y3
= 2y3
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
\(a,P=\left(5x^2-2xy+y^2\right)-\left(x^2+y^2\right)-\left(4x^2-5xy+1\right)\\ =5x^2-2xy+y^2-x^2-y^2-4x^2+5xy-1\\ =\left(5x^2-x^2-4x^2\right)+\left(y^2-y^2\right)+\left(-2xy+5xy\right)-1\\ =3xy-1\)
\(x+y=6,2\\ \Rightarrow y=6,2-1,2=5\)
Thay \(x=1,2;y=5\)
\(\Rightarrow3.5.1,2-1=17\)
`P = 5x^2 - x^2 - 4x^2 - 2xy + 5xy + y^2 - y^2 - 1`
`= 3xy - 1`
Thay `x = 1,2; y = 6,2 - 1,2 = 5` vào
`3 xx 1,2 xx 5-1 = 18 - 1 = 17`