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( mik k ghi đề nhé bn)
a) (2x)^3 - y^3 + (2x)^3 + y^3 - 16x^3 + 16xy = 16
=> 8x^3 - y^3 + 8x^3 + y^3 - 16x^3 + 16xy = 16
=> 16xy = 16
=> xy = 1
Vì x, y nguyên => x = 1, y = 1 hoặc x = -1, y = -1
mik xin lỗi nha, mik chỉ bt làm câu a
a) Ta có : (x - 5)2 - 16
= (x - 5)2 - 42
= (x - 5 - 4)(x - 5 + 4)
= (x - 1)(x - 9)
b) 25 - (3 - x)2
= 52 - (3 - x)2
= (5 - 3 + x)(5 + 3 - x)
= (x + 2)(8 - x)
c) (7x - 4)2 - (2x + 1)2
= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)
= (5x - 5)(9x - 3)
= 5(x - 1)3(3x - 1)
= 15(x - 1)(3x - 1)
\(B=\left(x-y-1\right)^2+3\left(y-2\right)^2+2005\text{ }\ge2005\)
\(C=\left(x^2+4x\right)^2-25\ge-25\)
\(2004.2006.\left(2005^2+1\right)=\left(2005-1\right)\left(2005+1\right)\left(2005^2+1\right)\)
\(=\left(2005^2-1\right)\left(2005^2+1\right)=2005^4-1< 2005^4\)
5xy2(x - 3y) = 5x2y2 - 15xy3
(x+5)(x2-2x+3) = x3-2x2+3x +5x2-10x+15= x3+3x2-7x+15
(x + 2y)(x-y)= x2-xy+2xy-2y2= x2+xy-2y2
tk mình nha bạn
Khó thế giải hộ tao bài này với Trang ơi:
( x + 2) ( 1 + x - x2 + x3 - x4 ) - ( 1 - x ) 9 1 + x + x2 + x3 + x4 )
a, x2-2x+1 b,9x2+6x+1
=x2-2x1+12 =(3x)2+2.3x.1+12
=(x+1)2 =(3x+1)2
c,x2+4xy+4y2
=x2+2x.2y+(2y)2
=(x+2y)2
d,49-14y+y2
=72-2.7y+y2
=(7-y)2
e,(x-y)2+2(x-y)+1
=(x-y)2+2(x-y).1+12
=[(x-y)+1]2
=(x-y+1)2
Chúc bạn học tốt!
\(a,x^2-2x+1=\left(x-1\right)^2\)
\(b,9x^2+6x+1=\left(3x+1\right)^2\)
\(c,x^2+4xy+4y^2=\left(x+2y\right)^2\)
\(d,49-14y+y^2=\left(7-y\right)^2\)
\(e,\left(x-y\right)^2+2\left(x-y\right)+1=\left(x-y+1\right)^2\)
Ta có:
\(x^2+y^2+5+2x-4y\)
\(=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)\)
\(=\left(x+1\right)^2+\left(y-2\right)^2\)\(>0\)
\(\Rightarrow\)\(\left|x^2+y^2+5+2x-4y\right|=\left(x+1\right)^2+\left(y-2\right)^2\)
\(-\left(x+y-1\right)^2\)\(< 0\)
\(\Rightarrow\)\(\left|-\left(x+y-1\right)^2\right|=\left(x+y-1\right)^2\)
\(\left|x^2+y^2+5+2x-4y\right|-\left|-\left(x+y-1\right)^2\right|+2xy\)
\(=\left(x+1\right)^2+\left(y-2\right)^2-\left(x+y-1\right)^2+2xy\)
\(=4x-2y+4\) (rút gọn nha)
\(=4.2^{2011}-2.16^{503}+4\)
\(=2^{2013}-2^{2013}+4=4\)
P/s: bn tham khảo nhé, mk ko biết đúng or sai, lm bừa
\(2x^2+y^2+2x-2xy+5-4y=0\)
\(\Leftrightarrow\left[y^2-2y\left(x+2\right)+\left(x+2\right)^2\right]+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(y-x-2\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-x-2=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
\(S=\left(x+2\right)^2+\left(y-1\right)^2=\left(1+2\right)^2+\left(3-1\right)^2\)
\(=3^2+2^2=13\)