Câu 1. Khai triển hằng đẳng thức(x-y)2 được kết quả là
A. x2 +xy+y2 B. x2 -xy+y2
C. x2 +2xy+y2 D. x2 -2xy+y2
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1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
Ta có:
x x - y - y y - x = x 2 - x y - y 2 - x y = x 2 - x y - y 2 + x y = x 2 - y 2
Chọn (B) x 2 - y 2
Lời giải:
a) (x2 + 2xy + y2) : (x + y)
= (x + y)2 : (x + y)
= x + y
b) (125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
c) (x2 – 2xy + y2) : (y – x)
= (x – y)2 : [-(x – y)]
= -(x – y)
= y – x
Hoặc (x2 – 2xy + y2) : (y – x)
= (y2 – 2yx + x2) : (y – x)
= (y – x)2 : (y – x)
= y – x
\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
b: (x-y)(x^2-2x+y)
\(=x^3-2x^2+xy-x^2y+2xy-y^2\)
\(=x^3-2x^2-x^2y+3xy-y^2\)
c: \(\left(x^2-y\right)\left(x+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y^2-xy-y^3-\left(x^3-y^3\right)\)
\(=x^2y^2-xy\)
d: \(3x\left(2xy-z\right)-5y\left(x^2-2\right)+3xz\)
\(=6x^2y-3xz-5x^2y+10y+3xz\)
\(=x^2y+10y\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
rút gọn P=2/x-(x2/(x2-xy)+(x2-y2)/xy-y2/(y2-xy)):(x2-xy+y2)/(x-y)
r tìm gt P với |2x-1|=1 ; |y+1|=1/2
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)
a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+\left(2y\right)^3-\left(x^3-y^3\right)\)
\(=x^3+8y^3-x^3+y^3\)
\(=9y^3\)
b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)
\(=x^3-x^2-x+1-x^3-8\)
\(=-x^2-x-7\)
\(\left(x-y\right)^2\)
\(=\left(x-y\right)\left(x-y\right)\)
\(=\left(x-y\right)x-\left(x-y\right)y\)
\(=\left(x^2-xy\right)-\left(xy-y^2\right)\)
\(=x^2-2xy+y^2\)
Vậy chọ đáp án D. \(x^2-2xy+y^2\)
Trả lời:
( x - y )2 = x2 - 2xy + y2
Vậy D là đáp án đúng.