Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]\)
\(=3\left(1-2xy\right)-2\left(1+3xy\right)\)
\(=3-6xy-2-6xy=-12xy+1\)
c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\)
\(=101^2-3\cdot101^2+3\cdot101+2012\)
=1002013
a) x4 + 3x3 - 7x2 - 27x - 18
= x4 + x3 + 2x3 + 2x2 - 9x2 - 9x - 18x - 18
= x3 . (x + 1) + 2x2 . (x + 1) - 9x . (x + 1) - 18(x + 1)
= (x + 1)(x3 + 2x2 - 9x - 18)
= (x + 1)[x2 .(x + 2) - 9.(x + 2)]
= (x + 1)(x + 2)(x2 - 32)
= (x + 1)(x + 2)(x + 3)(x - 3)
b) x4 + 3x3 + 3x2 + 3x + 2
= x4 + x3 + 2x3 + 2x2 + x2 + x + 2x + 2
= x3 (x + 1) + 2x2 . (x + 1) + x(x + 1) + 2(x + 1)
= (x + 1)(x3 + 2x2 + x + 2)
= (x + 1)[x2 .(x + 2) + (x + 2)]
= (x + 1)(x + 2)(x2 + 1)
\(x^4+3x^3-7x^2-27x-18\)
\(=\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(9x^2+9x\right)-\left(18x-18\right)\)
\(=x^3\left(x+1\right)+2x^2\left(x+1\right)-9x\left(x+1\right)-18\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+2x^2-9x-18\right)\)
\(=\left(x+1\right)\left[\left(x^3-3x^2\right)+\left(5x^2-15x\right)+\left(6x-18\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)+5x^2\left(x-3\right)+6\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)^2\)
b) \(2x^2+4y^2+z^2-4xy-2x-2z+5=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-2x+1\right)+\left(z^2-2z+1\right)+3=0\)
....
a) \(x^2+5y^2-4xy+6y+9=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y=2.\left(-3\right)=-6\\y=-3\end{matrix}\right.\)
Vậy : \(\left(x,y\right)=\left(-6,-3\right)\)
2x2 + 3y2 = 5xy
=> 2x2 + 3y2 - 5xy = 0
=> 2 ( x2 - 2xy + y2 ) - xy + y2 = 0
=> 2 ( x - y ) 2 - y ( x - y ) = 0
=> ( x - y )[ 2( x - y ) - y ] = 0
=> ( x- y ) ( 2x - 2y - y ) = 0
=> ( x - y ) ( 2x - 3y ) = 0
TH1 : x - y = 0
=> x = y
Thay x = y vào \(\frac{x+2y}{3x-y}\)
=> \(\frac{x+2y}{3x-y}=\frac{y+2y}{3y-y}\)\(=\frac{3y}{2y}=\frac{3}{2}\)
TH2 : 2x - 3y = 0
=> 2x = 3y
=> \(\frac{x}{y}=\frac{3}{2}\)
=> x = \(\frac{3}{2}.y\)
Thay x = \(\frac{3}{2}.y\)vào \(\frac{x+2y}{3x-y}\)
=> \(\frac{x+2y}{3x-y}=\frac{\frac{3}{2}.y+2y}{3.\frac{3}{2}y-y}\)\(=\frac{\frac{7}{2}.y}{\frac{7}{2}.y}=1\)
Câu 1:
a: Để M là số nguyên thì \(2x^3-6x^2+x-3-5⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
b: Để N là số nguyên thì \(3x^2+2x-3x-2+5⋮3x+2\)
\(\Leftrightarrow3x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-\dfrac{1}{3};-1;1;-\dfrac{7}{3}\right\}\)
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
Đang 3x2+3y2 sao lại ra -3(x+y)2 ?? Phải là +3(x2+y2) chứ :v Không nhớ hằng đẳng thức 1 và 3 à :v với cả 6xy đâu?