Cho x-y=1
Tính B=x^2-y^2-3xy
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Ta có:
\(x+y=1\)
\(\Rightarrow\left(x+y\right)^3=1^3\)
\(\Rightarrow\left(x+y\right)^3=1\)
\(\Rightarrow x^3+3x^2y+3xy^2+y^3=1\)
\(\Rightarrow x^3+3xy\left(x+y\right)+y^3=1\)
\(\Rightarrow x^3+3xy\cdot1+y^3=1\)
\(\Rightarrow x^3+3xy+y^3=1\)
Vậy: \(x^3+3xy+y^3=1\)
Có \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(x-\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)
\(\Leftrightarrow\left[x^2-\left(\sqrt{x^2+1}\right)^2\right]\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)
\(\Leftrightarrow-y-\sqrt{y^2+1}=x-\sqrt{x^2+1}\) (1)
Lại có:\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)\left(y-\sqrt{y^2+1}\right)=y-\sqrt{y^2+1}\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left[y^2-\left(\sqrt{y^2+1}\right)^2\right]=y-\sqrt{y^2+1}\)
\(\Leftrightarrow-x-\sqrt{x^2+1}=y-\sqrt{y^2+1}\) (2)
Từ (1) và (2) cộng vế với vế có:
\(-\left(y+x\right)-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)=x+y-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)\)
\(\Leftrightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\) hay S=0
Vậy...
\(x^2+y^2+z^2=1\Rightarrow x^2,y^2,z^2\le1\Rightarrow-1\le x,y,z\le1\)
Ta có:\(x^3+y^3+z^3-x^2-y^2-z^2=0\)
\(\Rightarrow x^2\left(x-1\right)+y^2\left(y-1\right)+z^2\left(z-1\right)=0\)
Vì \(x-1\le0,y-1\le0,z-1\le0\)
\(\Rightarrow x^2\left(x-1\right)\text{}\le0,y^2\left(y-1\right)\le0,z^2\left(z-1\right)\le0\)
\(\Rightarrow x^2\left(x-1\right)\text{}+y^2\left(y-1\right)+z^2\left(z-1\right)\le0\)
Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}x^2\left(x-1\right)=0\\y^2\left(y-1\right)=0\\z^2\left(z-1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left(x,y,z\right)\) là bộ (0,0,1) và các hoán vị
\(\Rightarrow x^{2021}+y^{2021}+z^{2021}=1\)
A=x^3 + y^3 + 3xy(x+y)
=x+3x^y+3xy^2+y^3
=(x+y)^3=2^3=8
B=x^2+2xy+y^2+4
=(x+y)^2+4=4+4=8
C=x^3+y^3+3xy(x+y)+7(x+y)
=(x+y)^3+7(x+y)
=2^3+7.2
=8+14=22
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
Lời giải:
$xy+\sqrt{(1+x^2)(1+y^2)}=1$
$\Leftrightarrow \sqrt{(1+x^2)(1+y^2)}=1-xy$
$\Rightarrow (1+x^2)(1+y^2)=(1-xy)^2$ (bp 2 vế)
$\Leftrightarrow x^2+y^2=-2xy$
$\Leftrightarrow (x+y)^2=0\Leftrightarrow x=-y$.
Khi đó:
$M=(x+\sqrt{1+(-x)^2})(-x+\sqrt{1+x^2})=(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x)$
$=1+x^2-x^2=1$
a) Ta có : \(\left(x+y\right)^3=1^3=1\)
\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)
\(\Leftrightarrow x^3+y^3+3xy=1\) ( do x + y = 1 )
a) \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(A=x^2+2x+y^2-2y-2xy+37\)
\(A=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(A=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(A=\left(x-y\right)^2+2\left(x-y\right)+1+36\)
\(A=\left(x-y+1\right)^2+36\)
Thay x - y = 7 vào A
\(A=\left(7+1\right)^2+36\)
\(A=8^2+36\)
\(A=64+36\)
\(A=100\)
b) \(B=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy-9\)
\(B=\left(x^3-3x^2y+3xy^2-y^3\right)+\left(x^2+xy-3xy+y^2\right)-9\)
\(B=\left(x-y\right)^3+\left(x^2-2xy+y^2\right)-9\)
\(B=\left(x-y\right)^3+\left(x-y\right)^2-9\)
Thay x - y = 7 vào B
\(B=7^3+7^2-9\)
\(B=343+49-9\)
\(B=383\)
c) \(C=x^3-x^2-y^3-y^2-3xy\left(x-y\right)+2xy\)
\(C=\left[x^3-y^3-3xy\left(x-y\right)\right]-\left(x^2-2xy+y^2\right)\)
\(C=\left(x-y\right)^3-\left(x-y\right)^2\)
Thay x - y = 7 vào C
\(C=7^3-7^2\)
\(C=343-49\)
\(C=294\)
d) \(D=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)-95\)
\(D=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy-95\)
\(D=\left(x^3-3x^2y+3xy^2-y^3\right)+\left(x^2-2xy+y^2\right)-95\)
\(D=\left(x-y\right)^3+\left(x-y\right)^2-95\)
Thay x - y = 7 vào D
\(D=7^3+7^2-95\)
\(D=343+49-95\)
\(D=297\)