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a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)
b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)
a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)
\(=x^6-9x^3+8-x^6+9x^3=8\)
b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)
\(=x^6-y^6-x^6+y^6=0\)
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) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
`# \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.
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
Lời giải:
$P=(x+1)^3-(x+1)^3-[(x-1)^2+(x+1)^2]$
$=-[(x-1)^2+(x+1)^2]=-[(x^2-2x+1)+(x^2+2x+1)]=-2(x^2+1)$ phụ thuộc vào giá trị của biến nhé. Bạn xem lại đề.
$Q=(2x)^3-y^3+(2x)^3+y^3-16x^3$
$=8x^3-y^3+8x^3+y^3-16x^3=(8x^3+8x^3-16x^3)+(-y^3+y^3)=0+0=0$ không phụ thuộc vào giá trị của biến (đpcm)
a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)
\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0