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Ta có \(x^2+y^2+xy+x=y-1\)
\(\Leftrightarrow2x^2+2y^2+2xy+2x-2y+2=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(\Rightarrow B=\left(-1+1-1\right)^{2023}\) \(=\left(-1\right)^{2023}\) \(=-1\)
Biến đổi: 4 x 2 − 4 xy + y 2 = 0 ⇔ ( 2 x − y ) 2 = 0 ⇔ 2 x = y
Thay y = 2x vào P ta được P = -3
B1
a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)
b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)
c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\)\(\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^3+1^3\right)\left(x^3-1^3\right)=x^6-1\)
d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)
\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)
\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)
\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)
B2:
\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)
\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)
\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)
Bài 1:
a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy+y^2\)
=4xy
b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)^2\)
\(=\left(2y\right)^2=4y^2\)
c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a+b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)
\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)
\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)
\(=2a^2-4bc\)
Tớ sẽ chứng minh đề sai:
\(\hept{\begin{cases}x+y=1\\xy=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2=1\\2xy=2\end{cases}}\Rightarrow x^2+4xy+y^2=3\) (Cộng theo vế)
Thay xy = 1 vào: \(x^2+y^2+4=3\Leftrightarrow x^2+y^2=-1\)
Mà \(x^2;y^2\ge0\forall x;y\)
Vậy tính A "=" niềm tin à? vì không có gì x,y nào thỏa mãn để tính cả!
Đề bài tương đương với \(2x^2+2y^2+2xy-2x+2y+2=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
\(M=x^{2020}+y^{2020}+2020^{x+y}=1^{2020}+\left(-1\right)^{2020}+2020^{1-1}=1+1+1=3\)
x2 + y2 + xy - x + y + 1 = 0
<=> 2( x2 + y2 + xy - x + y + 1 ) = 2.0
<=> 2x2 + 2y2 + 2xy - 2x + 2y + 2 = 0
<=> ( x2 + 2xy + y2 ) + ( x2 - 2x + 1 ) + ( y2 + 2y + 1 ) = 0
<=> ( x + y )2 + ( x - 1 )2 + ( y + 1 )2 = 0 (*)
Vì \(\hept{\begin{cases}\left(x+y\right)^2\\\left(x-1\right)^2\\\left(y+1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x+y=0\\x-1=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
=> x = 1 ; y = -1
Thế vào M ta được
M = 12020 + (-1)2020 + 20201-1
= 1 + 1 + 1
= 3