Cho x-y=-3 xy=10. Tính a. x^2- 2xy + y^2 -1 b. x^2+y^2
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a ) \(x^2-2xy+y^2-1\)
\(=\left(x-y\right)^2-1\)
\(=\left(-3\right)^2-1\)
\(=9-1\)
\(=8\)
b ) \(x^2+y^2\)
\(=x^2-20+y^2+20\)
\(=x^2-2.10+y^2+20\)
\(=x^2-2xy+y^2+20\)
\(=\left(x-y\right)^2+20\)
\(=\left(-3\right)^2+20\)
\(=29\)
a) \(x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)
b) Có: \(x^2-2xy+y^2=9\)
=> \(x^2+y^2=9+2xy=9+2\cdot10=9+20=29\)
Bài 2:
\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)
\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)
\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)
\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)
Bài 1:
a) \(A=x^2+2xy+y^2=\left(x+y\right)^2=\left(-1\right)^2=1\)
b) \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)
c) \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)
d) \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)
ta có A=x2-2xy+y2=x2+2xy+y2-4xy=(x+y)2-4 x (-10)=32+40=49
B=x2+y2=>B=x2+2xy+y2-2xy=(x+y)2-2 x (-10)=9+20=29
\(A=x^2+2xy+y^2-4xy\)
\(=\left(x+y\right)^2-4xy\)
= 49
\(B=x^2+2xy+y^2-2xy\)
\(=\left(x+y\right)^2-2xy\)
= 29
a: \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\)
=>\(\dfrac{xy}{5}=\dfrac{x^2+y^2}{8}=k\)
=>\(xy=5k;x^2+y^2=8k\)
\(A=\dfrac{8k-2\cdot5k}{8k+2\cdot5k}=\dfrac{-2}{18}=\dfrac{-1}{9}\)
b: Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\)
=>x=a*k; y=b*k; z=c*k
\(B=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)
\(=\dfrac{k^2\cdot\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
Bài 1:
Theo bài ra ta có:
\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)
\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)
\(=25-10y+y^2+25-10x+x^2-4\)
\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)
\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)
\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)
\(=50-50+5^2-4-4\)
\(=25-8=17\)
Vậy giá trị của \(\left(x-y\right)^2\)là 17
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
a, x2 - 2xy + y2 - 1
= (x - y)2 - 1
= (-3)2 - 1
= 9 - 1
= 8
b, x2 + y2
= x2 - 20 + y2 + 20
= x2 - 2.10 + y2 + 20
= x2 - 2xy + y2 + 20
= (x - y)2 + 20
= (-3)2 + 20
= 9 + 20
= 29