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\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
Ta thay : \(x=0;y=2009;z=2010\) ta được :
\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)
Chúc bạn học tốt !!!
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)
Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
\(A=\left(x+y\right)^2=\left(x-y\right)^2+4xy\)
Thay x - y = 5 và xy = 3 vào ta có:
\(5^2+4\cdot3=37\)
Vậy A = 37
B = \(\left(x-y\right)^2=\left(x+y\right)^2-4xy=7^2-4\cdot12=1\)
C sai đề?
D = \(x^2-y^2-2013x-2013y\)
\(=\left(x-y\right)\left(x+y\right)-2013\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(x+y\right)=0\)
E = \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x^2-y^2\right)\cdot0=0\)
bạn ơi cái biểu thức C= ( x + 2y ) mũ 2 biết 2y = x, xy = 8
mik xin lỗi nhé !!!
1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)
4. 4x2 + 4x + 1 = ( 2x + 1)2
5. \(\dfrac{1}{4}x-\dfrac{2}{3}xy+\dfrac{4}{9}y^2\) \(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.\dfrac{2}{3}+\left(\dfrac{2}{3}y\right)^2\)
\(=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)^2\)
6. \(4a^2-\dfrac{4}{3}ab+\dfrac{1}{9}b^2=\left(2a\right)^2-2.2a.\dfrac{1}{3}+\left(\dfrac{1}{3}b\right)^2=\left(2a-\dfrac{1}{3}b\right)^2\)
7.
\(9x^2+4xy+\dfrac{4}{9}y^2-25z^2=\left(3x+\dfrac{2}{3}y\right)^2-\left(5z\right)^2=\left(3x+\dfrac{2}{3}y-5z\right)\left(3x+\dfrac{2}{3}y+5z\right)\)