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\(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\)
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2 tháng 10 2021
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
20 tháng 4 2017
a) (x + 3)(x2 – 3x + 9) – (54 + x3) = (x + 3)(x2 – 3x + 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 – 2 . x . y + y2] – (2x – y)(2x)2 + 2 . x . y + y2]
= [(2x)3 + y3]- [(2x)3 - y3]
= (2x)3 + y3- (2x)3 + y3= 2y3
20 tháng 4 2017
Bài giải:
a) (x + 3)(x2 – 3x + 9) – (54 + x3) = (x + 3)(x2 – 3x + 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 – 2 . x . y + y2] – (2x – y)(2x)2 + 2 . x . y + y2]
= [(2x)3 + y3]- [(2x)3 - y3]
= (2x)3 + y3- (2x)3 + y3= 2y3
LS
1
DN
12 tháng 6 2018
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 10 2023
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)-3x^2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)+\left(x^2-y^2\right)-3x^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2-3x^2\)
\(=3x^2+y^2-3x^2\)
\(=y^2\)
5 tháng 8 2020
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
2 tháng 10 2020
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
\(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\)
\(=\left(x-y-x-y\right)^2-\left(2x\right)^2\)
\(=\left(-2y\right)^2-\left(2x^2\right)\)
\(=4y^2-4x^2\)
\(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\)
\(=\left[\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\right]-\left(2x\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2-\left(2x\right)^2\)
\(=\left(x-y-x-y\right)^2-\left(2x\right)^2\)
\(=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y\right)^2-\left(2x\right)^2\)